text:
Produce the number 2014 without any numbers in your source code

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It's 2017 2018 2019 2020 2021 2022 2023 2024 already, folks, go home.

Woo, 10 years of this challenge!

So, now that it's 2014, it's time for a code question involving the number 2014.
Your task is to make a program that prints the number 2014, without using any of the characters 0123456789 in your code, and independently of any external variables such as the date or time or a random seed.
The shortest code (counting in bytes) to do so in any language in which numbers are valid tokens wins.

Leaderboard:

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Even though numbers are ignored in brainfuck, I thought I'd post one anyway. 32 Chars: ++++++[>++++++++<-]>++.--.+.+++.

Brainfuck isn't a valid language for this challenge.

I know.  That's why I posted it as a comment

I wonder if this question gets a small spike in popularity around New Year's.

Waiting for "Come on folks, don't you realize it's 2016?" :)

@cagirici Done.

@BradenBest It's possible to do it in 31 characters in at least two different ways:
+++++++[>+++++++<-]>+.--.+.+++.

and
++++++++++[>+++++<-]>.--.+.+++.

@JoeZ. It's almost that time of the year again.

possible to do the brainfuck answer in 23 bytes -[>+<-----]>-.--.+.+++.

That language is full of language!

@JoeZ. We're discussing this question on meta.

C++:
cout<<to_string(('V'-'B'))+to_string(('V'-'H'))<<"\n";

(unsigned char)'\xca'*'\xa' for 2020 in C.

Locomotive basic (Amstrad CPC): ?ASC("-")*ASC("-")-&b, 22 bytes. (Cannot asnwer since the question is protected)

The question text says "any language in which numbers are valid tokens".  Since any character that is not an instruction is a comment in BF, numbers are valid therein.  Though admittedly you could go into a philosophical debate over whether digits count as tokens or whitespace in such a language.

50 bytes in brain-flak - ((((((()()()()()){}){})){}{}()){()()({}[()])}{}())

Mouse-2002, 4 bytes.
That's 4 bytes of pure, sweet ASCII.
In Mouse, the letters of the alphabet are initialised to the values 0-25. ! is the operator for printing integers, thus this prints 20 then 14 (no intermittent newline).
U!O!

There's no online interpreter available, but here you will find an interpreter written in C (needing some tweaks before one can coerce gcc to compile it) and the same compiled interpreter for Win32 but which works perfectly on Linux with wine.
Here you can find the fixed version of the interpreter, which compiles.

Wow, this is an amazing find so much later after the challenge. And with a language so esoteric nobody here had heard of it yet!

I just discovered this language today and I already love it. However, it's not purely esoteric, which makes it even better!

BTW: this does the same thing as CJam's KE, but is actually valid because it's a language from before 2014.

The smug look on my face right now resembles that of a Fortran-77 wizard after beating some proponent of Python in a [tag:fastest-code].

Also this solution will work until 2025 at least, until you pass U!Z!.

But by then, somebody will invent a language that prints out the Unicode value of any unrecognized token, and then the high score will be at 2 bytes forevermore.

This a game of @cat and Mouse-2002.

@DigitalTrauma indeed :3

Running the 2007 binary using Wine on Ubuntu, the program as written produces an error message to the standard output which doesn't include the string 2014. The modified program U!O!$ works, producing 2014 and no trailing newline.

@JoeZ, that's... that's kinda what I've done. ;)

Python 2, 51 bytes
print sum(ord(c) for c in 'Happy new year to you!')

Try it online!
Updated for 2015 thanks to @Frg:
print sum(ord(c) for c in 'A Happy New Year to You!')

Try it online!
Mouse over to see 2016 version:

 print sum(ord(c) for c in 'Happy New Year to you!!!')

Try it online!

You deserve the cleverness award

You forgot a print somewhere. Nice one otherwise :)

wow - this is so in the spirit of this question. <3

I was a little surprised that it was only the 4th phrase I tried after "Happy new year!", which would have been perfect for the year 1374.

Is this a rip-off of Mikael Mayer's answer in Scala, though?

No, I did not look at anyone else's answers first.

I only posted mine two hours before yours. This one is not that hard to find by oneself.

Ugh, ever since I started programming in Befunge, Python looks so wordy. But since this is such an amazing solution (+1), here is the equivalent Befunge: "Happy new year to you!"aa+k+.@

You want terse? Ruby: p 'Happy new year to you!'.sum

print sum(ord(c) for c in 'HAPPY NEW YEAR To you too.')
Oh wait, I'm a year late. That's 2013.

sum(map(ord,'Happy new year to you!')) would save 7 characters.

Even at my best I am a bogey golfer.

I created this account just to give +1!

sum(b'Happy new year to you!') in Python 3

sum(bytearray('Happy new year to you!')) in Python 2.6+

Awesome timing for awesome answer!

haha good answer,  in PHP : echo array_sum(array_map('ord',str_split('Happy new year to you!')));

You guys made my day! Thank you and have a great year wherever you are :)

Printable ASCII at 32-127 means about max 62 chars, min 15. You clocked 22. I'd like to see your list of things you considered before picking this one, added to the answer. If it's code golf, how much smaller while being literate could it get...margin is 7 chars.  :-)

It's hard to remember exactly, but the first thing I tried was "Happy new year!", then "Happy new year!!!!!", then "Happy new year everyone!", and finally 'Happy new year to you!'

Mind blown! That is very clever of you sir!

Updated: print sum(ord(c) for c in 'A Happy New Year to You!')

Uh....I get invalid syntax in 3.4? How does one run this?

For Python 3.x, print needs ( )'s around the expression.

Congrats on being the first answer on this site to reach 1,000 upvotes. :)

You don't need the space after ord(c)

print sum(ord(c) for c in 'Happy New Year to You!!!!') for 2017 :D

Why were revisions undone for print sum(map(ord,'A Happy New Year to You!'))? It does the same thing.

@SunnyPatel it's frowned upon to edit in golfs

When can haz 2017?

@Phoenix print sum(ord(c) for c in 'Happy New Year to You!!!!')

@dansalmo Perhaps you want to add that one to the answer as well? Would be cool to see the entire list in 2025. ;) (Oh, and I took the liberty to add a TIO-link.)

@KevinCruijssen That wasn't their solution. fedorqui posted that solution nearly a year prior, up a few comments.

Exactly 1234 upvotes at the time of the posting of this comment.

One byte can be saved by deleting whitespace before "for".

@stackzebra and also by deleting whitespace after in

I offer: print(ord('&')*ord('j')/len('NY'))

This does not make a serious attempt at solving the question (this is code-golf)

print(ord('ߞ')) for -36 bytes.

print(sum(ord(c) for c in 'have a happy new year!!'))

2018: print sum(ord(c) for c in 'Happy New Year To You!!!!!')

This answer is being discussed on Meta.

Go, 2 bytes (UTF-16)
One unicode character (2 bytes in UTF-16, 3 bytes in UTF-8 format), output 2014 as part of an error
—

http://ideone.com/dRgKfk
can't load package: package :
prog.go:1:1: illegal character U+2014 '—'

That's very clever, but the question requests a program that prints the number.

If printed error codes count, this submission should win.

+1. The question doesn't say, "prints only the number and nothing else".

I don't care if its valid or not, it's pretty awesome.

This is a standard loophole

I joined this community just to like this post.

@MilkyWay90 That's adding input, not adding output.

Ruby, 15
p Time.new.year

Temporary ;)

Note that the section of the question

independently of any external variables such as the date or time or a random seed

was not edited in until long after I posted my answer...

Jan Dvorak offers a great alternative in the comments:
Happy = Time
Happy.new.year

But it's so unenthusiastic. I prefer:
Happy = Time
class Time; alias year! year; end

Happy.new.year!

Or even:
class Have; def self.a; A.new; end; end
class A; def happy; Time; end; end
class Time; alias year! year; end

Have.a.happy.new.year!

And here's correct English punctuation:
def noop x = nil; end
alias a noop
alias happy noop
alias new noop
alias year! noop
def Have x
    p Time.new.year
end

Have a happy new year!

Okay okay, I couldn't help it:
def noop x = nil; end
eval %w[we wish you a merry christmas! christmas and a happy new].map{|x|"alias #{x} noop"}*"\n"
def year!; p Time.new.year; end

we wish you a merry christmas! we wish you a merry christmas!
we wish you a merry christmas and a happy new year!

Happy = Time; Happy.new.year

Befunge-98 (FBBI), 17 11 9 8 bytes
'-:*b-.@

Try it online!
Similar to the old version, but I remembered about '
'-:* pushes 45, duplicates it, then squares it, producing 2025
b-   subtracts 11 from it, resulting in 2014
.@   prints the result, then ends the program

Interestingly, \$45^2-11\$ is the only pairing of numbers a,b where
$$(a,b)∈[32,126]\times[10,15]\land a^2-b=2014$$
The significance of those sets is that \$[32,126]\$ is the set of printable ascii characters and \$[10,15]\$ is the set of easily accessible Befunge numbers. I found that pair with this python program:
for a in range(32,127):
    for c in range(10,16):
        if (a**2-c)==2014:
            print("%s,%s"%(a,c))

Or, if your interpreter supports unicode, then this works:
Befunge 98 - 5 bytes (4 chars)
'ߞ.@

It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars):
"ߞ".@

Befunge-98 (FBBI), 9 bytes
Old version:
cdd**e-.@

Try it online!
computes the number, then prints it:
cdd pushes numbers to the stack so that it is this: 12,13,13
**  multiplies top three values of stack, which is now: 2028
e   pushes 14
-   subtracts the top two values of the stack, resulting in: 2014
.   prints the numerical value
@   end of program

Befunge-98 (FBBI), 17 bytes
Older version:
"*'&("#;:a`j@a+,;

Try it online!
Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it.

http://en.wikipedia.org/wiki/Befunge ...what is this i don't even...

"The language was originally created by Chris Pressey in 1993 as an attempt to devise a language which is as hard to compile as possible"...

Wouldn't a:+.e.@ achieve the same? Saves a character by printing 20 and then 14.

@MSalters That would actually print 20 14

@Quincunx Where does it say that isn't allowed?

@11684 The output is rather well defined in the question. If a program printed: randomstuff2randomstuff0randomstuff1randomstuff4 I wouldn't consider it a valid solution.

I agree. But at least in my native language (Dutch) spaces can be used in numbers to make numbers more readable. So I wouldn't categorise a space as 'random stuff'.

@Plato Rather than read the wikipedia article (which only includes Befunge 93), if you want to learn about Befunge 98, read the official specs

By the same idea, '&:f+*.@ should work as well.

@MvG yeah, I was just manually searching for short solutions, then wrote a program, then tried different methods with that program. Of course there are other short methods.

Wiki: 'The language was originally created by Chris Pressey in 1993 as an attempt to devise a language which is as hard to compile as possible' ...but why?

Can I edit the question to include my answer?

Python 2, 26 bytes
print int('bbc',ord("\r"))

Try it online!

bbc is 2014 in base 13

Would've been my approach. Until I saw the accepted answer...

I assume the use of the characters 'bbc' and the use of base 13 is a Douglas Adams tribute. "I may be a sorry case, but I don't write jokes in base 13."

13 years ago. RIP.

IMO, this answer is in "True Pythonese"

Wikipedia reference for the Douglas Adams and base-13, http://en.wikipedia.org/wiki/Base_13

@BrianMinton: The use of Base 13 is probably because it's the only base between 2 and 36 that allows the number 2014 to be represented without any of the characters 0-9.

Coincidence? I think not! :-)

Woah I almost upvoted your bbc is 2014 in base 13 but then I realized it has 42 upvotes.  Perfection

Here is a permalink to Base 13 as of Jan 4 '14 for anyone interested. Be wary that it contains a mild spoiler to "The Restaurant at the End of the Universe".

MATLAB, Scala (4 characters, 5 bytes)
You can take advantage of MATLAB's (and Scala's) relatively weak type system, here.
The trick is to apply the unary + operation on a string composed only of the character ߞ (of UTF-8 code point U+07DE, or 2014 in decimal). This operation implicitly converts the string to a double (in MATLAB) and to an Int (in Scala):
+'ߞ'

Byte-count details:

+ is ASCII and counts for 1 byte
' is ASCII and counts for 1 byte (but appears twice in the expression)
ߞ is a 2-byte UTF-8 character

Total: 5 bytes
TeX (32 26 characters, as many bytes)
\def~{\the\catcode`}~}~\\~\%\bye

An even shorter alternative (proposed by Joseph Wright) is
\number`^^T\number`^^N\bye

XeTeX/LuaTeX (13 characters, 14 bytes)
If XeTeX or LuaTeX are allowed, UTF-8 input can be used directly (as proposed by Joseph Wright):
\number`ߞ\bye

\let~\number~`^^T~`^^N\bye 25 chars/bytes. You IMHO count wrong, it's 25 for your solution as well.

@tohecz I think both our solutions are actually 26-byte long.

If the current year is wanted, then 13 bytes: \the\year\bye

+'ߞ' also works in Scala (and a few other languages I imagine)

dc, 6 bytes
DiBBCp

Try it online!

D pushes 13 on the stack, even tho the input radix is 10 initially
i changes input radix (to 13 from 10)
BBC is 2014 base 13.
p prints.

Console output:
$ dc <<< "DiBBCp"
2014

dc<<<DiBBCp - a bit shorter

True, but the actual dc program is still DiBBCp (6 chars), the rest is just a way to run it.

I was going to upvote this but it has 42 points! uses base 13 and the word BBC. How cool is that! Seems that this year we will find the question for life, universe and everithing ;-)
http://en.wikipedia.org/wiki/Base_13.
I am upvoting @daniero's comment instead and leave the answer with this magnificent 42 reputation ;-)

@PabloMarin-Garcia, unfortunatelly some unaware person broke it... Has 43 votes now. Please go back and downvote! :)

@Tomas  Vogons always the Vogons. Resistance is futile against the intergalactic burocracy.

I cannot even find the D command in the man page. What does it do? Never mind... D is hex for 13.

I would upvote but you have 64 votes

@mpapis Or dc -eDiBBCp. Same length, I think.

Scala, 32 29 bytes
+"Happy new year to you!".sum

Try it online!
Well ok if you really want it golfed with any chars, you can use:
Scala, 11 bytes
'@'*' '-'"'

Try it online!
Scala, 22 bytes
"{yz}"map(_-'I'toChar)

Try it online!

Interesting idea, even if it's not particularly golfed. But the string itself makes it more interesting, than something using high bytes, or something like that.

I added a golfed version.

You could save a couple characters from the toInt like this: +"Happy new year to you!".sum Test

Updated. I did not know that + could also be a prefix.

R, 72 45 bytes
This is far from the shortest answer posted, but no one has yet posted an answer that

doesn't use character codes as a substitute for numbers, and
doesn't call the system date.

Using pure math (okay, and an automatic boolean conversion) in R, from the R console:
x<-(T+T);x+floor(exp(pi)^x)*x*x-(x*x)^(x*x)/x

Try it online!
Prints out the number 2014.  T is a pre-defined synonym for true in R.  The floor and exp functions are directly available in the base package, as is the pi constant.  R doesn't have an increment operator, but repeating the (x*x) turned out to be fewer characters that doing increment and decrement twice each.

Original version in Javascript (72 characters)
For the simple reason that I could test out in the console, and it doesn't mind a complete lack of whitespace:
m=Math;p=m.pow;t=true;++t+m.floor(p(m.exp(m.PI),t))*t*t++-p(++t,t--)/--t

run in your console and it will print back the number 2014.

Props to xkcd (and also) for getting me to think about exp(pi):

P.S.  If you can make the same algorithm shorter in a different language, post a comment with it.

o=!"";(o+o)+""+(o-o)+(o+o-o)+(o+o+o+o)

25 bytes: alert('ߞ'.charCodeAt())

+1. However, you claim your answer was the first answer that "doesn't use character codes as a substitute for numbers, and doesn't call the system date". That is actually false. My answer has this solution cdd**e-.@ (posted before yours) which does not make use of character codes or system date. It computes the number 2014. c,d, and e are hexadecimal number digits. a,b,...,f push (respectively) 10,11,...15 so 12 * 13 * 13 - 14 is the computation.

This wasn't the first answer to not use character codes; I have several earlier ones.

@Quincunx: you're right; skimming through I didn't catch that your version was using hex digits, not character codes.

o=+!'';t=o+o;''+t+(+!o)+o+(t+t), a bit smaller than @kernel-james one

R code with 38 characters x=T+T;y=x*x;x+floor(exp(pi)^x)*y-y^y/x

Nice catch, @djhurio.

I also like @KernelJames version (and Ricardo's improvement) though they really qualify as a completely different method rather than an improvement on mine -- I assume the "protected" status wasn't letting you post?

two k implementations of this algo, both 39 chars: 1: -_-(x*p*p:exp acos@-#`)-f*x*x*x:f*f:#`` ; 2: f+(x*_p*p:exp acos@-#`)-f*x*x*x:f*f:#`` . I can't seem to squeeze it any further at the moment....

and here it is in APL in 31 chars (which i'm quite new to, so i'm sure it can be improved--particularly the hacky generation of the 2 constant): ⌈(x×p×p)-f×x×x×x←f×f←⍴⍕⌊p←*○~⊃⍬

C (clang), 33 bytes
main(){printf("%d",'A'*' '-'B');}

Try it online!

How annoying that the prime factorization of 2014 requires ASCII 5!

I got  for(int a;YES;a++){NSLog(@"%i",a);}  for Cocoa Touch objective C but I can't add yet (not rated enough). It does show a 2014 eventually and it has a compiler error and is possible it may not work at all and compiled is probably about 4.2 meg - but hey.

main(){printf("%d",'\a\xde');}

@mjy Byte order of multi-character literals is not guaranteed.

@JonathonReinhart multi-character literals are entirely implementation-defind.

On specified env. using multichar is good

PHP, 9 bytes
This requires PHP 7.1 or lower. It will work in PHP 7.2, but it will result in a warning. No guarantees for any future version.
xxd needed because of binary data (so copying and pasting it would be easier). May return E_NOTICE, but it doesn't really matter, does it?
~ $ xxd -r > 2014.php
0000000: 3c3f 3d7e cdcf cecb 3b                   <?=~....;
~ $ php 2014.php
2014

Alternatively, save this using ISO-8859-1 encoding.
<?=~ÍÏÎË;

@Quincunx: Uploading this was relatively tricky, but here is Ideone - http://ideone.com/APHKhn.

This looks interesting; what does it do?

@Quincunx: It does bitwise not (0x00 changes into 0xFF, and 0x7F changes into 0x80) on every character of the string. As the string is valid identifier (anything with high bit set is an identifier character for PHP, probably to support other encodings), PHP thinks it's a constant, but because it's not defined, it treats it as a string.

Since the question was "Now that it is 2014...", how about echo date('Y');?

@John: Nope. If you would read the comments for the question, you would notice that OP doesn't want the to get current year as an answer. Besides, <?=date(Y); would be 11 characters, and I have solution in 9 characters.

Why not <?=~ÍÏÎË instead?

@IsmaelMiguel: I guess (if you save it using ISO-8859-1).

Python3.4.0b2 (0 bytes)
% python3.4
Python 3.4.0b2 (v3.4.0b2:ba32913eb13e, Jan  5 2014, 11:02:52)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>>

This won't work once people update their version of Python 3 a year from now.

That's why I mentioned python3.4.0b2, It wont happen twice. Its release date, not system date.

This made me smile!

Also works in Python 3.4.2.

You are not allowed to add extraneous output.

Mathematica, 14 characters (or 15 if you count the bitmap as a character)
TextRecognize@

Actually, you should count bytes so it's 14 + the size of the bitmap image.

The bitmap would probably be 7,357 characters, really.

Also, I believe some of the bytes in the image fall into the \d range. Namely, if it's the GIF, then the header itself is guaranteed to contain some.

You are all killjoys. Fine, use First@ToCharacterCode@"ߞ" for 25 characters, 26 bytes.

@MichaelStern Yup, that's been done allready.

@Ajasja only after I fixed it. The old solution as given produced a list of answers rather than '2014' as specified by the problem.

As a bonus, here's a Mathematica solution in 30 characters --URLFetch@"http://goo.gl/miVwHe"

I liked this answer so much I made it into its own question: http://codegolf.stackexchange.com/q/17588/2605

JavaScript (Node.js), 23 bytes
Uses Base 64 Conversion
alert(atob("MjAxNA=="))

Try it online!
Or
alert("MMXIV") // ;)

MMXIV has no characters in common with 2014, last I checked... // ;)

The ancient Romans might disagree.  The first one produces 2014 though, in modern English.

Your first one doesn't output though...

if you open the console in chrome, paste it and hit enter it returns the value - I'll add an alert to it as its longer than others anyway.

@logic8: Nice one +1. Another JavaScript version which also works: (4 bytes) [!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]] - alert([!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]]); Though I'm not sure it meets Joe Z's requirements.

@FrançoisWahl, Now that's cool!  Nice one.

Or go the other way btoa("ÛMx") for 11 characters

Just so you know, atob("MjAxNA==") returns "2014" (a string). +atob("MjAxNA==") returns a number. Doesn't matter for alert, though.

@logic8 The ancient Romans would probably find it quirky as they didn't use generally use subtractive notation (e.g. IV for 4).

Here's another javascript base one: alert(parseInt('bbc',+[]+'xd'))

@PeterCooper: That's interesting. Would the standard have been "IIII" then?

@logic8 You can omit the equals signs to get 21 chars

alert("ߞ".charCodeAt())

Perl - 10 characters
This solution is courtesy of BrowserUK on PerlMonks, though I've shaved off some unnecessary punctuation and whitespace from the solution he posted. It's a bitwise "not" on a four character binary string.
say~"ÍÏÎË"

The characters displayed above represent the binary octets cd:cf:ce:cb, and are how they appear in ISO-8859-1 and ISO-8859-15.
Here's the entire script in hex, plus an example running it:
$ hexcat ~/tmp/ten-stroke.pl
73:61:79:7e:22:cd:cf:ce:cb:22
$ perl -M5.010 ~/tmp/ten-stroke.pl
2014

Perl (without high bits) - 14 characters
say'````'^RPQT

This uses a bitwise "or" on the two four-character strings "RPQT" and "````" (that is, four backticks).
$ ~/tmp/fourteen-stroke.pl
73:61:79:27:60:60:60:60:27:5e:52:50:51:54
$ perl -M5.010 ~/tmp/fourteen-stroke.pl
2014

(I initially had the two strings the other way around, which required whitespace between print and RPQT to separate the tokens. @DomHastings pointed out that by switching them around I could save a character.)
Perl (cheating) - 8 characters
This is probably not within the spirit of the competition, but hdb on PerlMonks has pointed out that Perl provides a variable called $0 that contains the name of the current program being executed. If we're allowed to name the file containing the script "2014", then $0 will be equal to 2014. $0 contains a digit, so we can't use it directly, but ${...} containing an expression that evaluates to 0 will be OK; for example:
say${$|}

For consistency, let's do the hexcat-then-perl thing with that:
$ hexcat 2014
73:61:79:24:7b:24:7c:7d
$ perl -M5.010 2014
2014

I think this is cheating, but it's an interesting solution nonetheless, so worth mentioning.

Hey there, if you reverse the string and barewords, you can save a char: print"````"^RPQT. It might be possible to use say too on 5.10+ using -E instead of -e, but I don't know if that incurs a +2 penalty for different command-line args?

It's considered acceptable to specify that you're using Perl 5 and use say for no penalty.

@DomHastings, personally I think if you're going to do it with "-e" or "-E", I'd say that you should then have to include the entire command, including "perl -E" in your character count.

@PeterTaylor, say is a syntax error in Perl 5 unless you explicitly write a pragma to enable it such as use v5.10 or use feature 'say'. In Perl 6 it's enabled automatically.

Relevant meta answer

You may also save another byte by using string literals that don't require quotes, such as say ABCD^srrp.

Ruby, 20 bytes
p 'bbc'.to_i ?\r.ord

Try it online!
Explanation: bbc is 2014 in base 13. Shorter than Python. Not as short as Forth.

"Nobody writes jokes in base 13!"

w=?$;"^XA[_AXeMFGIAHJLjKNAEFEJJNHQHNKLAEMINJOJOHLAGKHOJOJ[AG[HQHRFJAH}IHAIGGwIIAHHGwKHAHGHrEUAGQFiGVAGQGfIPAFHKHHbJHAQII]MGASHNSOHATIdIAUJJRLIAWLIQGKAZOFUA]ZAeSAiPAjOAkLA".codepoints{|r|r-=68;$><<(r<0??\n:(w=w==?$?' ':?$)*r)}

p 'ߞ'.ord for a fully functional program, 'ߞ'.ord inside irb. Works on 1.9+

Without "strange" unicode characters: ?-*?--?\v (45 * 45 - 11) (Ruby 1.8)

One character smaller: ?.*?.-?f (46 * 46 - 102) (Ruby 1.8 only again)

How is the base 13 indicated in the print?

@RecycledSteel, to_i takes a base parameter.  I'm passing it 13 since that's the ordinal value of a carriage return (\r).

z=''.to_i;o=Math.cos(z).to_i;t=o+o;"#{t}#{z}#{o}#{t+t}" without relying on charcodes

@RicardoPanaggio, those are some sweet answers. You should submit. I'm going to leave mine as-is, simply for the BBC - Base-13 - Douglas Adams connections. :-)

(late comment) Can you do i? instead of i ??

PowerShell, 9 bytes
+"ߞ"[""]

Try it online!
ߞ (U+07DE NKO LETTER KA) is counted as two bytes according to the code-golf tag info.
[""] returns the first character from the string ("" is converted to 0). The unary plus opeartor (+) converts the character to an integer.

According to http://codegolf.stackexchange.com/tags/code-golf/info, it should count as 9 characters.

@GlitchMr, thanks, I've corrected my answer.

The [''] is a nice trick. I usually used [char] but that's indeed shorter :)

Javascript, 18 characters

alert(btoa('ÛMx'))

Update: in ES6, using a template literal can save two characters:

alert(btoa`ÛMx`)

The code above is fairly easy to understand by keeping in mind that btoa converts a string into another string according to a set of well-defined rules (RFC 4648). To see how the conversion works, we're going to write the input string "ÛMx" as a sequence of binary digits, where each character is rendered as its 8-bit character code.

Input character          |        Û |        M |        x
Character code (decimal) |      219 |       77 |      120
Character code (binary)  | 11011011 | 01001101 | 01111000

After reorganizing the binary digits in the last row in groups of 6, we get the binary representation of 4 new numbers, corresponding to the Base64 indices of the 4 characters in the string "2014".

Base64 index (binary)  | 110110 | 110100 | 110101 | 111000
Base64 index (decimal) |     54 |     52 |     53 |     56
Output character       |      2 |      0 |      1 |      4

As per HTML specification, the output characters can be retrieved from their Base64 indices according to this table: http://dev.w3.org/html5/spec-LC/webappapis.html#base64-table.
If you don't care about the details, you could let the browser do the calculations for you and find out that "ÛMx" is the result of evaluating atob('2014') in Javascript.

the question mentions printing the number. I know that evaluation in a console will lead to the result being returned, but on [CodeGolf.SE] it's generally expected that JS answers will include some form of IO (alert is most common).

@zzzzBov Yes, my original answer included an alert call.

It lists Community as having removed the alert. I wonder if the standards have changed for JS codegolf or whether it's just a change that was adopted by Community.

@zzzzBov: It was a change by anonymous user accepted by SHiNKiROU and luser droog. Anyway, this edit triggered a discussion on Meta (http://meta.codegolf.stackexchange.com/questions/803/javascript-standards-for-io).

@xfix, It's not this edit alone that caused me to ask that question on meta, I've seen enough code golfs where people get into disagreements over whether the dev console counts as printing, that I figured it was worth asking for a standard.

I'd love to have an explanation of how this works.

@DA. I added a short explanation.

Or in a function expression (allowed by default) x=>btoa\ÛMx``

@programmer5000 Times have changed but the consensus in 2014 was to use alert. You can find the discussion in the link in xfix's comment.

if functions are allowed instead of full program: _=>btoa'ÛMx' which is 13 bytes (the ' symbols should be `, but i can't because of the limitations in comments)

Scala, 6 bytes
"?="##

Try it online!
(## is Scala's symbol meaning hashCode, and the Java string "?=" hashes to 2014.)
Scala, 5 bytes
+'ߞ'

Try it online!
Math on our favorite unicode character produces an Int.

GolfScript, 9 bytes
Yet another GolfScript entry. I believe this is shorter than any of the printable GolfScript entries so far:
"!="{*}*)

Try it online!
(Peter Taylor's 7-char entry beats it, but includes non-printable control characters.)
I call this the "that's so last year!" entry, because what it actually does is generate the number 2013 in 8 chars, as 33 × 61, and then increments it by one. ;-)

You win (for ascii golfscript)

C (gcc), 31 bytes
main(){printf("%o",' b'/'\b');}

Try it online!
C (gcc), 32 bytes
main(){printf("%x%o",' ','\f');}

Try it online!
C (gcc), 30 bytes
main(){printf("%x",' j'^'~');}

Try it online!
C (gcc), 30 bytes
main(){printf("%d",'\a\xde');}

Try it online!

How about just "%x" and 'b'*'S' ? That's 8212 dec or 2014 hex. Saves one char one literal and two on format string.

@MSalters No, 'b'*'S' is 8134. The prime factorization of 8212 is 222053, so I don't see an easy way to produce it.

Bah, checked with integer division that 8212/'b' == 'S' :(

and octal doesn't work either (02014=1036=37*28, 28 is unprintable)

@MSalters The octal was a nice hint though, see my new 31 solution ;)

@FredOverflow what about my 28 char solution?
main(){puts(DATE+'\a');}

@ArunprasadRajkumar Will it work in the next year too? (eg. display "2014")

Your last solutions employs implementation defined behavior and is likely invalid.

The most recent edit included a variation of code which breaks the rule of including numbers in the code. It uses the number "5".

28 characters: main(){printf("%o\b",' a');}  A space is 32 in decimal. 'a' is 97. Appending them in a character literal, 32*256+97=8289. 8289 decimal is 20141 in octal. Following that printout with a backspace produces just the required 2014.

27 bytes

GolfScript, 14 bytes
'-+,/'{)))))}%

Try it online!
How it works: ASCII goes like this:
...
+
,
-
.
/
0
1
2
3
4
...

So, this takes the ASCII codes of each character, subtracts five, and sticks it in a string.
{...}% yields an array of the characters of a string mapped, when given a string as an argument. So, it increments each character by 5 () means increment).

The + at the end is unnecessary. Also, rather than increment 5 times, just add five. Total savings: 4 chars. '-+,/'{5+}%

@Quincunx, 5 is a number.

@PeterTaylor I keep forgetting.

Clever solution

The CJam version of this: "-+,/"_,)f+. :) (Then again, CJam has KD.)

Forth (gforth), 14 bytes
'> '" * '^ - .

Try it online!

Can you explain how this works?

I'm expressing integer constants as character literals using their ordinal (ASCII) values. So this is: 62 34 * 94 - .  If you don't speak Forth, this means print (62 * 34 - 94).

Forth (Gforth 0.7.0), 5 bytes, 4 chars: 'ߞ . It prints the character.

Python 2, 32 bytes
print ord(',')*ord('-')+ord('"')

Try it online!
Probably possible to reduce it using the 2014 th Unicode char ߞ, but I didn't try.
Quincunx notes that
a=ord('.');print a*a-ord('f')

Try it online!
is shorter by three chars.

The unicode version works, at least in Python 3 with IDLE: print(ord("ߞ")). It does not appear to work in Python 2; Python 2 probably does not support unicode.

Also, with your version, a=ord('.');print(a*a-ord('f')) is 2 chars shorter.

@Quincunx It does support, however you need to have a specific encoding set to your file and append u at the end of the string (would be "ߞ"u)

@Kroltan Actually, I'm pretty sure the second part is not true. In Python 2, you had to prepend strings with u to say that they were unicode, but in Python 3, all strings are automatically unicode

@murgatroid99 Ah, sorry. It's been a while since I used python, and my brain remembered about C-style float literals (1453f)

Similar as commented above, without unicode print ord('.')**2-ord('f') is slightly shorter.

APL (6 bytes, 4 chars)
⊃⎕TS

Only works this year though.
Why it works:
      ⎕TS
2014 1 1 11 58 5 811
      ⊃⎕TS
2014

Without relying on the system date, it's 10 bytes (7 characters):
⎕UCS'ߞ'

Unfortunately, answers that only work for this year are invalid.

I don't know APL:  is that square box an APL char, or am I missing a font representation (Chrome) ?

@JoeZ.: The question doesn't say that but I've added one that doesn't rely on it being 2014.

@CarlWitthoft: It's called a quad (⎕), it's supposed to look like that.

Python 2, 32 bytes
Had some fun writing this:
my_lst = []
for i in range(33, 126):
    for j in range(i, 126):
        if 2014 - 126 < i * j < 2014 - 33:
            if j not in range(48, 58):
                my_lst.append("ord('" + unichr(i) +
                              "')*ord('" + unichr(j) +
                              "')+ord('" + unichr(2014 - i * j) + "')")

for val in my_lst:
    print val, '->', eval(val)

Prints all the possible ways I can write 2014 using Bruno Le Floch's method:
ord('!')*ord(':')+ord('d') -> 2014
ord('!')*ord(';')+ord('C') -> 2014
ord('!')*ord('<')+ord('"') -> 2014
ord('"')*ord(':')+ord('*') -> 2014
ord(')')*ord('/')+ord('W') -> 2014
ord('*')*ord('-')+ord('|') -> 2014
ord('*')*ord('.')+ord('R') -> 2014
ord('*')*ord('/')+ord('(') -> 2014
ord('+')*ord(',')+ord('z') -> 2014
ord('+')*ord('-')+ord('O') -> 2014
ord('+')*ord('.')+ord('$') -> 2014
ord(',')*ord(',')+ord('N') -> 2014
ord(',')*ord('-')+ord('"') -> 2014

Python 2, 10 bytes
But this is obviously redundant, so if your interpreter is set to utf-8 by default, then all it takes is:
ord(u'ߞ')

Try it online!
Python 2, 83 bytes
Also, thanks to AmeliaBR (for the idea), I tried my best to implement a pure math version:
from math import*
a,b,c=int(e),round(e),ceil(pi);print int(a**(b*c-(c-b))-a*a**c-a)

Try it online!

ord(',')**2+ord('N')

@MLS Well, that has a digit in it :)

I optimized your math version a bit: a,b=int(e),int(pi);c=a+a;print a**(b*c-c+b)-a*a**c-a, and you can eliminate the math import altogether by making use of the fact that True in Python 2.x is identical to the integer 1 in operation, bringing it down to 50 characters: o=True;a=o+o;b=a+o;c=b+o;print a**(b*c-o)-a*a**c-a

You can save a character if you use Python 3: ord('ߞ')

@JoachimIsaksson A digit is no number ;)

Here is an optimized math version (42 characters) a=True;a+=a;print(a<<a*a*a)*a*a-(a<<a*a|a)

R, 20 bytes
@popojan (he is not allowed to post an answer here yet) has provided the solution within 20 characters.
sum(T+T:exp(T+pi))-T

Try it online!
Output:
[1] 2014

R, 22 bytes
Anonymous user has suggested shorter solution.
strtoi("bbc",pi*pi+pi)

Try it online!
2014 is BBC in base 13. pi*pi+pi (=13.0112) is treated by R in this context as the integer 13. Output:
2014

R, 30 bytes
cat(a<-T+T,T-T,T/T,a*a,sep="")

Try it online!
Output:
2014

R, 31 bytes
cat(T+T,T-T,T/T,T+T+T+T,sep="")

Try it online!
Inspired from the answer by AmeliaBR. Output:
2014

Wow, +1! And I hoped my roman numerals solution will be the best... :( ;-)

The variants below can be made shorter: cat(a<-T+T,T-T,T/T,a^a,sep="") (30 chars), paste0(aT+T,T-T,T/T,a^a) (26 chars)

@Tomas, the function paste0 is not allowed, as it contains the symbol 0. Your solution makes 30 characters if using cat.

create file 2014 in current directory and write list.files() - 12 characters

@Zbynek, I expect your solution not to be valid in the general scope of codegolf rules.

For 2024, 26 characters, using the fact that 2024 is a tetraedric number : sum(cumsum(T:exp(pi)-T))

Java (JDK), 77 75 bytes
75 characters if print is added in a class with the main method:
class C{public static void main(String[]a){System.out.print('#'*'<'-'V');}}

Try it online!
It means 35*60-86 which is equal to 2014

I like your formula.

This actually needs to be longer, because the class definition needs to be in it to run. Shortest way I know (with Java 6):class a{static{System.out.println('#'*'<'-'V');}}

@Quincunx, with Java 7, a main method is needed

@TrueSoft Note I said "with Java 6". You should choose the shortest Java possible.

You can remove 'ln' in println to reduce chars...

@Quincunx, a class without main() would throw an error at runtime, and I don't think it's ok to put that as an answer.

Another fun way would have been value of "?=".hashCode(). Not shorter, just fun.

@TrueSoft see http://codegolf.stackexchange.com/a/16099/9498

I can improve this answer (to 72 chars) but I can't find the answer button on this StackExchange.. weird :S

You can shave 4 bytes off the expression if you're willing to use non-printables: 'j'*'' (106*19).

Read it and weep: class C{public static void main(String[]a){System.out.print(0xFBCA/' ');}}

My solution (http://codegolf.stackexchange.com/a/18051/14156) comes at 69 chars. Beats @HotLicks' solution by 5 chars :)

@JAnderton - Of course, that's a non-ASCII solution.  Different category.

@HotLicks 0x has a digit.

@Boann - Drat!!!  Forgot about that.

CJam, 2 bytes
KE

Try it online!
K and E are variables preset to 20 and 14.
I created CJam in 2014 so it's ok if it doesn't qualify.

GolfScript, 10 8 7 chars
This solution contains non-printable characters. As xxd output:
0000000: 2714 0e27 7b7d 2f                        '..'{}/

As base 64:
JxQOJ3t9Lw==

Unpacks the ASCII codes for characters 20 and 14 and prints those numbers without any separation.
To actually generate the number 2014, I'm currently at 8 chars:
0000000: 2713 6a27 7b2a 7d2a                      '.j'{*}*

JxNqJ3sqfSo=

Takes a string containing characters with ASCII values 19 and 106 and multiplies them to get 2014.

Does it really take 3 characters to output the ascii codes for a string? That's sad :/

Printing 20 and 14 without separation is acceptable.

I guess this can be actually shortened to 6 chars with {xy}.* where x and y represent 20 and 14 in ASCII.

@Vereos, I somehow missed seeing your comment earlier. Very nice, and sufficiently different that it's worth posting as a separate answer.

Alright, I will :)

Haskell, 23 characters, 24 bytes
main=print$fromEnum 'ߞ'

That's U+07DE N'Ko letter KA

According to http://codegolf.stackexchange.com/tags/code-golf/info, it has 24 characters.

Methinks that's 23 characters, @GlitchMr, though it may be 24 bytes.

@TRiG: Well, yes. The info page says that "If you use Unicode, byte count should use UTF-8."

Mathematica, 47 bytes
Total@ToCharacterCode@"Happy New Year for all!"

2014

If you remove the space between All and !, you can make All lowercase.

@JoeZ. Yes!.. tks

That's pretty awesome.

EXCEL, 24 characters:
=COLUMN(T:T)&COLUMN(N:N)

How does this even work?

@Timtech T:T and N:N define an entire column, and COLUMN returns the column number of the reference passed in (of its top left cell, IIRC). & is string concatenation.

Good job there :)

How about the sheet references to shorten it even more?

Why not =COLUMN(T1)&COLUMN(N1)?

@Neil because that has digits.

you can shorten to 16 bytes by using =COLUMN(BYL:BYL)

Mathematica, 23 bytes
Tr@ToCharacterCode@"ߞ"

(Using the 2014 unicode char. Credit for Tr goes to alephalpha)

Mathematica, 46 bytes (pure math)
a=⌊E⌋;b=⌈E⌉;c=⌈Pi⌉;a^(b*c-(c-b))-a*a^c-a

idea taken from here.

That returns a list. To return only the desired string without using any numbers, try First@ToCharacterCode@"ߞ". That's 25 characters, 26 bytes.

@MichaelStern Hmm I would think a List is OK.

Nope, has to be the number itself.

@MichaelStern Tr@ToCharacterCode@"ߞ".

@MichaelStern Thanks, great idea! JoeZ: Oh well:)

2 bytes "=[LongDash]" where "[LongDash]" is replaced to character and wolfram alpha returns 2014.

I and others have posted shorter Mathematica solutions to this challenge, but here are two methods that I don't believe have been tried in any of the other examples, in any languages.
Mathematica, 27 characters / 27 bytes
FromDigits["MMXIV", "Roman"]

Mathematica, 37 characters / 37 bytes
(ToString /@ FromDigits /@ {"K", "E"}) <> ""

and the following 58 character variation
FromDigits[Flatten[IntegerDigits /@ FromDigits /@ {"K", "E"}]]

37 chars:(ToString /@ FromDigits /@ {"K", "E"}) <> ""  By the way, why does FromDigits["K"] return 20??

@DavidCarraher Ah, 37 characters indeed, thank you. As for FromDigits["K"], "0"->0, "1"->1, "2"->2, "3"->3, "4"->4, "5"->5, "6"->6, "7"->7, "8"->8, "9"->9, "A"->10 etc.

Mathematica (24)
Found by randomly generating expression trees. TraditionalForm chops 5 characters for the floor symbol but I think rich text doesn't count.
Floor[π E(Sin@E+E^E)E^E]

+1 Very nice! Pi counts as two chars (takes two bytes in UTF8, so it's 25 bytes). You can get it to 22 bytes by ⌊π*E(Sin@E+E^E)E^E⌋ (as you mentioned).

D
ϯ has the numerical value 1007 in Unicode encoding.
as compiler message during compilation (20 chars)
pragma(msg,'ϯ'+'ϯ');

runtime version (45 chars)
import std.stdio;void main(){write('ϯ'+'ϯ');}

Couldn't you improve this by ditching the import and just qualifiying the write call (e.g., void main(){std.stdio.write('ϯ'+'ϯ');})

No, the compiler still requires the import.

(late comment) Can you do pragma(msg,'ߞ'); and import std.stdio;void main(){write('ߞ');}? NOTE: ߞ IS U+07DE (2014).

Arithmetical results are always widened to at least 32-bit integers, which is why the two character values add up and print as "2014". A single character retains its type and should be printed verbatim.

J - 11 bytes - Base arithmetic
+:_bj*p:_bf

Explanation
_bf and _bj are 15 and 19 in infinite base, therefore p:_bf is 15th prime, which is 53. The result is then 19*53 doubled, which is 2014.
J - 31 28 bytes - Math
(((!>:&+:)-(!>:))+:&+:)>.^*_

Explanation
The implementaion of A004126 sequence from OEIS where I searched for the ways to calculate 2014. Basically it's C(2*n+1,3)-C(n+1,3) where n=12.
This means ceiling(exp(signum(infinity))), which is 3
>.^*_

To get 12 we double it two times.
+:&+:

Everything else is a combo of a dyadic fork and monadic hook. You can read about them here.
Other's solution in J
+/(>.^*_)&u:'Happy new year to you!'

Which normally is +/3&u:'Happy new year to you!', it's a shame we need a number to work with strings.
It was fun to remember J :).

it's a shame we need a number to work with strings. You don't. You could do +/a.i.'Happy new year to you!'

Another 11 char J solution: *./a.I.'&j'.

Java (68)
Solution 4: 68 chars (Thanks @radiodef))
class C{public static void main(String[]a){System.out.print(+'ߞ');}}

Solution 3: 69 chars (Thanks @Kamran)
class C{public static void main(String[]a){System.out.print(~~'ߞ');}}

Solution 2: 71 chars
class C{public static void main(String[]a){System.out.print('޾'+' ');}}

Solution 1: 72 chars
class C{public static void main(String[]a){System.out.print((int)'ߞ');}}

<3 Unicode.

Solution Derivation Technique: Copy the output from the following line and convert that character back to int for your a working solution.
System.out.println((char)2014));

I'm pretty sure that 1 is a digit and hence forbidden.

Mine is better, without any numbers, and has 69 character.
class C{public static void main(String[]a){System.out.print(~~'ߞ');}}

Didn't think of that. Nice solution @Kamran

By the way instead of the double complement, you can basically cheat and use the useless "positive" operator. System.out.print(+'ߞ'); All it will do is promote to int.

Note:  This current answer produces text, not a Number object, as the question requested.

@Radiodef Updated the answer! Thanks for the suggestion.

@djangofan not sure if @JoeZ. meant it to be a number specifically. Either way, I don't think your statement is completely true. The only way Java can bit compliment the value is to consider it an Integer (or int). Here's a SSCCE that proves my point: Object o = ~~'ߞ'; System.out.println(o.getClass()); the output to which is class java.lang.Integer

@PeterTaylor Thanks for the note. I removed that solution. The current solution, you'll find, is compliant to requirements (to the best of my knowledge ;))

Tip: you can use enum C{ instead of Class C{ Some IDEs think its a compiler error  but it compiles flawlessly. that can result in another byte shaved off

n->+'ߞ' for 8 bytes (7 characters), since lambdas are appropriate equivalents of a program.

@OlivierGrégoire In that case you no longer need the +. Simply o->'ߞ' will be enough, if the return-type is an int. Try it online. But the challenge states "Your task is to make a program", so I would just leave it as is. Maybe changing class C{public static void main to interface C(static void main to save a byte, by going from Java 7 to 8.

Perl 6, 37 bytes
This is not my actual solution (the real solution is in PHP, and is shorter), but I decided to make shorter version of Python's solution, because it's a really interesting idea.
say [+] ords 'Happy new year to you!'

Try it online!
So, what does it do? Well, it gets list of character positions (ords function), sums them together ([] is a reduce operator, which takes the exact operator between square brackets, in this case +), and say prints the result with new line. Could be WAY shorter by abusing ~^ prefix operator (which isn't implemented in Rakudo Perl) or Unicode characters, but this is just for fun solution.
Perl 6 is a quite an interesting language, even if it's not as good as Perl 5 in most code golf tasks (because of mandatory whitespace in many situations, and generally less DWIM). However, in this case, because of builtin sum and ords, it wins with Perl 5.

At first I thought you meant your solution was in Perl and was 6 characters long. And then I was like, "no, that can't be right."

say ord "ߞ" is shorter but certainly less whimsical.

JavaScript 59, 44, 36, 29
a=+'';b=''+a+++a++;a+++b+ ++a

Assuming alert can be skipped due to implicit return

a=+'';b=''+a+++a++;alert(a+++b+ ++a)

alert([-~-~'',+'',-~'',-~-~-~-~''].join(''))

Those are l characters not 1
l='length';alert([[,,][l],[][l],[,][l],[,,,,][l]].join(''))

this is nice. +1

Python 3, 9 bytes
license()

Try it online!
unfortunately this does not work on tio.run but it does work on my device
license() stdouts text which luckily contains 2014 :)
...
2. Subject to the terms and conditions of this License Agreement, PSF hereby
grants Licensee a nonexclusive, royalty-free, world-wide license to reproduce,
analyze, test, perform and/or display publicly, prepare derivative works,
distribute, and otherwise use Python alone or in any derivative version,
provided, however, that PSF's License Agreement and PSF's notice of copyright,
i.e., "Copyright (c) 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010,
2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022 Python Software Foundation;

All Rights Reserved" are retained in Python alone or in any derivative version
prepared by Licensee.
...

This particularly only works on IDLE.

@UndoneStudios, rm main.py;echo "license()" >> main.py;python3 main.py and python3 -c "license()" works

I can’t tell if this breaks the loophole of outputting a long stream of numbers and saying “it’s in there somewhere!”

AutoHotkey 31/13
send % ++(z:=true) z-z true z+z

This year special solution
send % A_year

27 bytes: a:=Asc("")Asc(""), Send %a%. The characters in the Asc functions don't show correctly SE for me but they're the ASCII characters 20 and 14 copied and pasted into AHK. They show up as DC4 and SO (or maybe S0) in SciTE.

Golfscript - 11 10 chars
'&~I'{}/-*

computes 38 * 53 (ie an expansion of the prime factorization of 2014: 2 * 19 * 53)

Old Version
'f.'{}/.*\-

This computes 462-102.
'f.'{}/ puts the ascii values of f and . on the stack.
.*      squares the value of the .
\-      swaps the two values, then subtracts

Interestingly, the Befunge version of this Golfscript ('&~I'{}/-*), "&~I"-*.@, is one char longer than my Befunge answer and one character shorter than Golfscript.

Note that 2*9=18; what you meant was 2*19=38.

GolfScript, 10 bytes
I can't beat Quincunx's 10-char plain ascii solution but I can match it:
'N,,'{}/*+

Try it online!
(it calculates 44*44+78)
GolfScript, 26 bytes
And now a different approach :)
'codegolfing is yummy'{+}*

Try it online!

Yet another 10-char solution: "xr"{`(;}% (free bonus smiley included!)

Raku (Perl 6), 12 bytes
say 'ߞ'.ord

R using roman numbers, 29 characters
as.numeric(as.roman('MMXIV'))

@plannapus yeah I know crazy idea :) Well, but quite normal in ancient times :)

probably a "legacy" function :)

Nice! There are 28 characters in as.double(as.roman('MMXIV')).

C#, 60 characters
class a{static void Main(){System.Console.Write(- -'ߞ');}}

Thanks to mleise for the original answer for D!
Note: space between hyphens is important

1 is not allowed

I came up with 'ϯ'+'ϯ' but I believe this is 1 char shorter.

if an exit code is acceptable class a{static int Main(){return - -'ߞ';}} is just 42 chars.

No need for "- -'ߞ'", could just do +'ߞ'.  So: class a{static void Main(){System.Console.Write(+'ߞ');}} which is 56 bytes

If we want to get without using the ASCII characters this is simple,
int i;
i = "  ".Length;
Console.WriteLine(i + "" + ((i-i)) + "" + --i + "" + ++i*i);

System.Console.Write(DateTime.Now.Year) has the added advantage that it will solve similar future contests as well :)

@JuliaHayward But it will fail this contest next year...

@JuliaHayward and it will also fail this contest this year... (Reread the OP).

OK, tongue-in-cheek comment failed utterly

VS 2010 crashes when I try this with a fresh new Console Application...

Ruby — 8 characters
p ?ߞ.ord

where ߞ is Unicode character U+07DE — N'Ko letter ka.
This is what it looks like when I copy and paste the above line and run it in my Terminal. Notice the character does not display:

I understand what you're trying to do, but help me out... How do I test this?  Copy & pasting your code doesn't work.  If I copy & paste the little "07DE" Unicode char then I actually see \U+FFDF in irb, but that char has an ord of 85.

I tested it by copying the symbol from the site listed above and running it in pry. I don't think Stack Exchange recognizes the symbol by default (so simple copy & paste of the above won't work), but Ruby 2.0+ should.

This script prints nothing.

How did you enter it? It works for me. Also, what version of Ruby are you using?

Interesting. The N'Ko alphabet reads right to left. Interspersing it with Roman characters seems to shuffle characters depending on where you edit/paste. Has anyone else been able to get this to run, or am I the only one?

p ?ߞ.ord would meet the requirement of printing.

this works for me

@steenslag I must have been snoozing to miss the print requirement. I thought returning 2014 was sufficient. Corrected. Thanks for pointing that out!

Python 3 (17 bytes)
print(ord("ߞ"))

Python 2 (50 bytes)
s,t=str,True;print s(t+t)+s(t-t)+s(t/t)+s(-~t<<t)

Python 2 version can be shortened to t=True;print't+t'+'t-t'+'t/t'+'-~t<<t', where the ' have to be changed to backticks. (I don't know how the enter real backticks that are shown as such).

@WolframH wrap the code-block in an extra backtick. Like so: t=True;print`t+t`+`t-t`+`t/t`+`-~t<<t`  where the initialization and end of the code block are done by use of: \`. Alternatively, replace every `in the code with\: `t=True;print\`t+t\`+\`t-t\`+\`t/t\`+\`-~t<<t\

Those are actually 16 and 49 bytes respectively. The trailing newline doesn't count.

JavaScript:
alert(+!![]+!![]+[+[]]+ +!![]+(+!![]+!![]+!![]+!![]));

each +!![] is 1 (![] is false, !false is true and +true is 1)
+[] is 0, and [0] is "0"
so this turns into 1+1+"0"+1+(1+1+1+1)

guess this is already beaten but I like the approach:
<?php
$three=round(pi());
$one=pi()/pi();
$two=$three-$one;
$four=$three+$one;
$zero=$one-$one;
echo $two.$zero.$one.$four;
?>

T-SQL, 50 40
My original entry seems too straightforward.  How's this?
PRINT RIGHT(CHECKSUM(',,.'),LEN('````'))

(Disclaimer:  Given Steve Matthews's comment, the above may be dependent on configuration.)
Here's the SELECT version in SQL Server 2012.
Original entry:
This may look obscene in more ways than one, but it's valid.
DECLARE @ INT=\PRINT CONCAT(-~-~@,@,-~@,-~-~-~-~@)

Try it in SQL Server 2012.  Here is a version using SELECT.
Explanation:  "\" is a valid currency symbol.  (Do a find for "String to" here.)  If you assign just a currency symbol to an INT variable, you store 0.  Also, "@" is a valid variable name.  "~" is bitwise NOT, and "-" is negative.  If you negate a NOT-ed INT, you get the INT plus 1.  So, repeat "-~" until you make @ into the digit you need.  Then CONCAT() your digits.

Running PRINT RIGHT(CHECKSUM(',,.'),LEN('````')) on MS SQL Server 2008 R2 produces the response "1200"

Can I suggest instead something like PRINT ASCII('')*ASCII('j') which is just 27 bytes (note that the character that isn't rendered here is the DC3 CHAR(19) in the first set of quote marks.

@Steve I actually ran the checksum one on the same version, too, but you touch on a good point that maybe the checksum algorithm could vary based on configuration.  I should add a disclaimer.  Also, I like your better 27-byte solution.  You should post it as an answer so people can vote it up.

PRINT UNICODE(N'ߢ') has 20 bytes

@JanDrozen Very nice, but you must've meant 'ߞ'.

Bash 28
echo $((`printf %d \':`#yG))

Similar to the Ruby and Python solutions, but uses base62, now without literal 6 and 2 (but pretty long...)! Using base58 as recommended by @DigitalTrauma along with removing quotes around the printf format string, Thanks!
Better 17 (Thanks @user13955 and @DigitalTrauma!)
tr a-l /-:<<<dbcf

"It's shorthand for tr abcdefghijkl /0123456789:. Not very geeky, I'm afraid." -user13955
Also updated to use a heredoc instead of echo, thanks @DigitalTrauma!

Not a valid solution since 2 and 6 are forbidden.

haha, Oh dear...

user13955 suggested this edit, might be worth considering: echo dbcf|tr a-l /-:
It's shorthand for tr abcdefghijkl /0123456789:. Not very geeky, I'm afraid.

I have a solution in pure bash/ksh. i.e. it does not use any external program like tr.  However, I lack the 10 reputation to post this. ;-)

Pure bash|ksh 110 bytes: sum(){ (IFS=+;echo $(($*)));}&&d(){ echo $((n=n+n));}&&false||n=$?&&sum $(d;d;d;d;echo $((n=n+n+n+n));d;d;d;d)

I verified that the unicode solution works in ksh93.

As the question doesn't seem to preempt showing any other characters in addition to 2014, how about 'date' for 4 characters? HNY

@DavidDelMonte, I think that answer has been suggested a few times, but it has to be 2014 specifically, otherwise, you'd be on to a winner!

Use a bash here-string to shave 3 chars of the tr solution: tr a-l /-:<<<dbcf

And no need for the quotes around printf format string: echo $((\printf %d "'>"`#wu))`

And use base 58 and drop one more char: echo $((\printf %d ':`#yG))`

@DigitalTrauma Amazing! Updated, thank you!

@DigitalTrauma Pure bash 18: echo $[$[$#xd]#bbc] ... without fork!

@F.Hauri Good one!

PHP: 54 chars
Slightly different approach, albeit long, using concatenation, and not using ordinal values:
<?=floor(M_E).floor(M_EULER).ceil(M_EULER).ceil(M_PI);

@SurabhilSergy that is not a valid answer, the output will change in 2015.

Javascript (29 chars)
t=!'',a=t+t+'';a+!t*t+t*t+a*a

Not the shortest but some boolean logic

That was my first idea too haha. This is the best I could get: a=!'';a+a+""+(+!a)+(+a)+(a+a+a+a) (33 chars)

@Benno (+!a) is pretty clever, didn't know it works like this

Yeah, it seems to treat it as an integer, only when in brackets though! Without the brackets, it reverts to 2falsetruetruetruetruetrue ;)

PHP - 13 bytes
<?=m_n_^_o_k;

A fairly trivial ASCII-only solution.
Other variants:
<?=Q__Q^cone;
<?=VonV^d__b;
<?=PAAP^bqpd;
<?=BEAU^pupa;

etc.

BEAU^pupa LOL.

<?=date('Y');
in any year )

@DmitryDubovitsky <?=date(Y); would also work, but it won't always print 2014, as the problem requires.

Javascript, 37 bytes
((_=+!![])<<''+_+''+_)-(_<<_+++_+_)-_

It's not code golf until someone bitshifts.
$ node
> ((_=+!![])<<''+_+''+_)-(_<<_+++_+_)-_
2014

It is actually Javascript, I promise.
((_=+!![])    // create 1 and save in _
  <<          // left bitshift by 11 to get 2048
    ''+_+''+_ // create 11 from '1' + '1'
  )
  -           // subtract (2048 - 32) === 2016
  (
    _<<       // bitshift by 5 to get 32
  _+++_+_     // create 5 from 1 + (++1) + 2
)-_           // subtract 2 to get 2014

That looks so much like Brainfuck.

Japt, 3 bytes (2 chars)
Japt is newer than this competition (created in 2015), but was not created specifically to answer it.
#ߞ

That's right. 2 chars. Explanation:
#   // Char-code of next character
 ߞ  // Character with char-code of 2014
    // Implicit output

Try it online!

Does this mean I can undelete my Microscript answer?

@SuperJedi224 There's a list in the main post of "invalid but interesting answers". I think newer languages fall into this category.

C, 33 30 bytes
Numerical Solution 1
f(){printf("%d",'<'*'#'-'V');}

// 2014 = 60 * 35 - 86

C, 86 83 bytes
Numerical Solution 2
#define A ((int)'}')
#define B (((int)'r')-((int)'d'))
f(){printf("%d",A*B+A+A+B);}

// A = 125
// B = 14
// 2014 = 125*14 + 125 + 125 + 14

C, 53 50 bytes
ASCII Art 1
f(){printf("┌┐ ┌┐ ┐ ┐┌\n┌┘ ││ │ └┤\n└┘ └┘ ┴  ┴");}

Result
┌┐ ┌┐ ┐ ┐┌
┌┘ ││ │ └┤
└┘ └┘ ┴  ┴

C, 94 91 bytes
ASCII Art 2
f(){printf("╔═╗ ╔═╗ ╗ ╦ ╦\n  ║ ║ ║ ║ ║ ║\n╔═╝ ║ ║ ║ ╚═╣\n║   ║ ║ ║   ║\n╚═╝ ╚═╝ ╩   ╩\n");}

Result
╔═╗ ╔═╗ ╗ ╦ ╦
  ║ ║ ║ ║ ║ ║
╔═╝ ║ ║ ║ ╚═╣
║   ║ ║ ║   ║
╚═╝ ╚═╝ ╩   ╩

You could shorten all your mains to f, saving 3 bytes each, and your submission will still be valid.

Vim 7.4, 18 bytes
:h os_v<CR>$bbbyiwZZp

Here, <CR> means a literal newline. Uses the help page os_vms.txt which contains the number 2014 on the first line.
How it works:
:h os_v<CR>            This opens the help page for os_vms.txt, the line the cursor lands on contains '2014'.
           $bbb        Move the cursor to the '2014' (end-back-back-back)
               yiw     Copy it (yank-in-word)
                  ZZ   Close the help page
                    p  Paste the copied word

If numbers were allowed we could do:
i2014

It's 2010 for me instead

@KritixiLithos Weird, it works for me. What help page does :h os_v open for you?

It opens os_vms.txt says Vim version 7.3 and last change was made in 2010 Jul 8

What version of Vim are you using?

I'm using Vim version 7.3

Perhaps it is the difference in OS (I'm using macOS El Capitan)

@KritixiLithos I'm using 7.4 (also Capitan), that's probably why. I'll update the answer accordingly.

If numbers were allowed you could just do i2014. :P

@JoeZ. Oh right. I'm stupid :P

TI-BASIC, 12 bytes
A completely new approach.
iPart(π+iPart(e)sinh(π+√(π

This could probably be golfed further, using a 1-byte token in place of e, but I have spent far too long on this today...

Javascript, 6 characters (8 bytes)
I never saw any rule saying we had to produce the number 2014 in the absence of any other output.
(Nor anything about not outputting to an error, but that's more obvious.)
new`—`

For me, on Firefox Nightly, this produces TypeError: "\u2014" is not a constructor, which contains the number 2014.
(If it isn't obvious, this happens because the em dash, —, is U+2014. Or, in other terms, it's the unicode character that can be represented by the hexadecimal number 2014.)

Yes, but... Simply — yield Uncaught SyntaxError: illegal character U+2014 on my Firefox.

Javascript: 47 characters
Not the shortest, but another solution:
$ node
> [!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]]
"2014"

Source
Javascript: 6 characters

Your task is to make a program that prints the number 2014 without using any of the characters 0123456789

Ok, here you go: 6 characters.
$ node
> "²⁰¹⁴"
'²⁰¹⁴'

Yeah, only for ASCII lowers. :-)
Will I win? :-D

Ruby:
p eval("#{""=~//}x#{"\a".ord}de")

Here is how it works inside IRB shell:
>> p eval("#{""=~//}x#{"\a".ord}de")
2014

TI-Basic, 49 41 31
π/π+π/π→B:B^B^B→C:C^C(C/B)-CB-B

Explanation

π/π+π/π→B Store 2 in B
:B^B^B→C Store 16 in C
:C^C(C/B)-CB-B Display 2014

I'm up voting on the assumption this works; I have no way to test it.

@MichaelStern Explanation added :)

:π/π+π/π→B:B^B^B→C:Disp C^C*(C/B)-C*B-B would be 2 characters shorter.

@MichaelMadsen Thanks, updated!

8 bytes can be saved with various tweaks: iPart(e→B:BBBB:Ans^B(Ans/B)-AnsB-B

Another 23-byter: iPart(e→B:BBBB^(BBB)-BBBBB-B

That can be 21 without assignment to B, using Ans instead. I would advise at least fixing your current solution, as it currently outputs a number well over 2014. That should be C^B instead of C^C.

J (18)
This one doesn't use any character codes. Uses the idea that the sum of the sequence of natural numbers from 1 -> 63 = 2016.
<:<:+/i.*:+~+~p:%_

You could save two bytes <:<:+/i.*:+~dyad

Multiboot x86 Assembly (250 bytes):
[BITS N]
%define L(x) mov dword [eax],'zaza'-x
O equ 'A'<<`\xe`
F equ 'bab'-'aaa'
M equ `\xFF\xAF\xAA\xFA`-`\xFD\xFF\xFC\xDE`
C equ -(M+F)
h:
dd M
dd F
dd C
dd h+O
dd h+O
dq 'a'-'a'
dd s+O
s:
mov eax,'\'<<`\r`
L('HBJB')
add eax,'e'-'a'
L('IBFB')
hlt

Compile with:
nasm src.S -o 2014.bin -f bin -DN=32

(If someone has a suggestion for replacing the N macro, I'm all ears)

You can also submit the machine code as its own answer, provided final compiled binary doesn't have any bytes from 0x30 to 0x39 in it.

x86 - 16 bytes (Assembled)
This little snippet moves the stack so it overlaps the video memory, to then push the individual numbers straight to the screen (They even blink! Surely a sign of good times to come).
Code: (Note: '0'-'9' -> 0x30-0x39)
B890A8 8ED0 2C5C 50 2C04 50 48 50 40 40 50

Assembled from:
mov ax, 0xA890
mov ss, ax
sub al, 0x5c
push ax
sub al, 3
push ax
dec ax
push ax
inc ax
inc ax
push ax

Build'n run:
nasm -f bin -o 2014.com 2014.asm
dosbox 2014.com

I think it shouldn't have any characters [0..9] in your code

@TeunPronk but that's the generation code, not the submission -- the submission is the binary

Golfscript, 6 chars
{xy}.*

Inspired from Peter Taylor's answer.
This solution involves non-printable characters too.
Where x and y represent 20 and 14 in ASCII (the non-printable characters we were talking about before).

awk (28)
There's definitely a need for an "awky" answer... ;-)
BEGIN{print++I+I--I++I++I*I}

...oookaaayyy... the last * may be a + too. But please don't call it an "awkf*ck" solution then... ;-)
BEGIN{print++I+I--I++I++I+I}

I think, I prefer the later version now because of less different characters...
(tested with gawk and mawk)

A different use of variables that saves one byte: BEGIN{print++I+I J++I++I*I}

Windows Calculator - 5 characters
Inspired by this answer to a different question.
Open the Windows Calculator in Programmer View (Hex mode) and type:
DFEC±
This is what the result looks like.

If you change the size to Word, it will be an almost-valid answer that will output only 2014

I don't think this is a programming language...

BASIC v2.0 (Commodore 64), 24 bytes
PRINT ASC("&")*(ASC("V")-ASC("!"))

The Commodore BASIC tokens are single-byte representations of BASIC keywords. This should result in storing the above line as 24 bytes instead of 34.
I was aiming to construct an equation based on character codes, so I was looking for ideal characters in PETSCII. To minimize the number of operations in the equation, it seemed like a good idea picking a pair of divisors of 2014.
However, none of the combinations were perfect, because either one of the divisors were too large, too small or the character code of a number.
So I ended up using 38 which is CHR$("&") and 53, but instead of directly using the latter which is CHR$("5"), I used the difference of 86 and 33. These are the character codes for CHR$("V") and CHR$("!") respectively.
Tested in VICE and FC64.

Emotinomicon, 14 bytes
ߞ

Try it online!
Explanation:
  ߞ       explanation
              begin quote string
    ߞ
              end quote string
              pop N and output as a number

ߞ is U+7DE. 0x7DE is 2014.

Okay,  thanks! In your code, do you convert the quote string to a number when it's outputting?

@MilkyWay90 Emotinomicon pushes the character code of each character in the string.

The Shakespeare Programming Language, 219 bytes
I am using drsam94's compiler.
.
Ajax,.
Puck,.
Act I:
Scene I:
[Enter Ajax and Puck]
Ajax:
You is the difference between a fat fat fat fat fat fat fat fat fat fat fat cat and the sum of a fat fat fat fat fat cat and a fat cat!Open thy heart!
[Exeunt]

Explanation
.

Everything from the first line to the first period is the title, parsed as a comment.
Ajax,.
Puck,.

These are characters from Shakespeare's plays, and descriptions (also comments)
Act I:
Scene I:

Act I and Scene I, used for gotos (not used here)
Ajax:

The character speaking.
You is the difference between

Assigns the difference of the next two values to the character being spoken to.
a fat fat fat fat fat fat fat fat fat fat fat cat

Every adjective multiplies by 2, a noun is either 1 or -1 depending on its connotation (in this case it's 1). 2^11 * 1 = 2048.
the sum of a fat fat fat fat fat cat and a fat cat

The sum of the next two values. 32 + 2 = 34, so 2048 - 34 = 2014
Open thy heart!

Prints the value of the character being spoken to as a number.

><>, 7 4 bytes
"nߞ

Try it online!
Explanation
"nߞ      : Put the string nߞ onto the stack.            Stack: [110, 2014]
 n       : Print the top item of the stack as a number. Stack: [110]
  ߞ      : Error out.

Huh, this worked really well in Runic too. One byte shorter than the solution I came up with a couple weeks ago. Not sure why I didn't try that before.

05AB1E, 6 bytes
T·žvÍ«

Uses the CP-1252 encoding. Try it online!
Explanation:
T       # Push 10
 ·      # Multiply by 2
  žv    # Push 16
    Í   # Subtract 2
     «  # Concatenate
        # Implicit output

+1 from me. Here a 6-byte alternative: Txs4+J (Push 10; Duplicate and double copy (10,20); swap; Add 4; Join together).

5-byter 'ߞÇ. Uses default ASCII encoding (so both ߞ and Ç are 2 bytes each), because ߞ doesn't exist in 05AB1E's codepage.

PHP, 27 bytes
not in any way competitive to xfix´s solution, but here are a few versions with 27 bytes each:
<?=hexdec($c=hexdec(e)),$c;   # e->14->20, e->14
<?=ord(U)-ord(A),hexdec(e);   # 85-65,14
<?=($c=ord(","))*$c+ord(N);   # 44*44+78
<?=($c=ord("."))*$c-ord(f);   # 46*46-102

The first one already has been posted by brother Filip; but the others have not.

This one has 37 bytes, but I like it (inspired by the Calculator solution)
<?=dechex(hexdec(dfeb)^hexdec(ffff));

or 13 bytes with
<?=sqpu^AAAA;

But that idea has already been used by primo

Rockstar, 34 32 bytes
X was up equalizing a word
Say X

Say hello to rockstar! (No I didn't make this one)
Explanation:
This first line:
X was up equalizing a word

Uses Rockstar's Poetic number literals. (As opposed to regular literals, which use numbers).
This means that the length of every word after was indicates the digit in that position. So here we have a 2 length (up), a 10 length (equalizing) a 1 length (a) and a 4 length (word). The length is % 10, so equalizing becomes a 0, and the result is that the variable X has the value 2014.
Then of course we print it with Say

Save 2 bytes by using X's instead of X was.

05AB1E, 4 bytes
ToRÀ

Try it online!
T        # 10
 o       # 2**x (1024)
  R      # reverse (4201)
   À     # rotate left (2014)
         # implicit output

Boring alternative: ŽGMR.

Some more: ŽFÌ;, ŽG_À, ŽΓû¨, ŽlT¦

Add++ -i, 13 bytes
L,"j"O"&"OL/*

Try it online!
Caird why no halve built-in you have like 3 fricking built-ins for doing special stuff with doubling but absolutely nothing for halving why do you do this to me?
Explained
L,"j"O"&"OL/*
L,             # start a lambda that's automatically called by the -i flag
  "j"O         #    push the number 106 to the stack - that's the ascii value of j
      "&"O     #    push 38 to the stack - that's the ascii value of &
          L/   #    push the length of the stack (2) and divide 38 by that. This is not a definitive way to consistently divide by 2...it's a miracle the circumstances allowed this. This results in 19
            *  #    and multiply the two items (106 and 19) to get 2014.

C++, 50 bytes
#include<iostream>
int main(){std::cout<<'U'^'A';}

You should provide compile-ready code in C++.

Also, use four spaces before each line for blocks of code.

i added "iostrean but it disappeared, actually am new so don't know how to"

This is supposed to be code golf, i.e. shortest code possible. Remove unnecessary whitespace, etc

@DoorknobofSnow thanx, after your edition i can still see few extra spaces

Rather verbose. int('U'-'A') prints 20, int('O'-'A') 14. Even shorter: ~~'U'-'A' also is 20. Result: #include <iostream> int main(){std::cout<<~~'U'-'A'<<~~'O'-'A';}.

Even more shorter: 'U'^'A'.

@MSalters Did you try compiling this? g++ on all standard versions gives an error: no match for ‘operator^’.

Doesn't work!

LiveScript, 18 bytes
The temporary solution
new Date!.getYear!

Unicode
\ߞ .charCodeAt!

Over Excitement
x=!Happy
Happy = -> console.log it
New = -> +it
Year = ->++x and Year
Year.valueOf = -> x

Happy New Year!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I initially assumed that you mean LiveScript, as in, JavaScript in first Netscape 2 beta.

@GlitchMr, that's where the name came from :-)

Only the "Unicode" solution is valid. The "temporary" solution violates the rule "... independently of any external variables such as the date or time"

... and the "Over Excitement" solution is not a serious contender,

C, 37 Characters
main(){printf("%d%d",'d'-'P','\xE');}

You could knock off 3 bytes by renaming main to f since we don't require main to be used.

MATLAB (no char codes), 47 chars
p=pi,q=p^p;e=exp(p);floor(q*q+q*e-q-q-q-p-e-e)

I played with a similar approach in Mathematica (which has a much larger set of mathematical constants built-in). How long did it take you to develop that solution.

Just playing around, maybe five minutes, probably less. I’m pretty sure this is far from the most compact formula.

JSFuck, 1267 bytes
In Javascript, here is the alert(2014) !
(Try in browser Console).
[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]][([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]((![]+[])[+!+[]]+(![]+[])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]+(!![]+[])[+[]]+(![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]]+[!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]])()

This sample uses only six different characters to write and execute code. This was generated by https://github.com/aemkei/jsfuck.

Even though it's the longest answer rather than the shortest, +1 for JSFuck.

Python 2, 68 bytes
Not really short, but it abuses the fact that Python accepts booleans where an integer is required, because bool is a subclass of int.
import string;d=string.digits;T=True;print d[T+T]+d[:T+T]+d[T+T+T+T]

+1 for the first non-cheating program in this question that doesn't depend on ASCII or UTF-8 (unless I missed something)

False can be shortened to d>d. On the other hand, d[d>d]+d[T] can be shortened to d[:T+T], so that optimization is not really necessary ;-). Also, from string import*;d=digits;saves one character.

@WolframH Noted. I'll update it later.

bash 35
I want something that doesn't use ASCII or UNICODE.
[ ];x=$?;y=$[x+x];echo $y$?$x$[y+y]

If you don't mind some error messages mixed in (28 characters):
[;x=$?;echo $x$?$[x/x]$[x+x]

You can get rid of the error messages if you don't mind corrupting a file (32 characters):
[ 2>x;x=$?;echo $x$?$[x/x]$[x+x]

That doesn't use ASCII? That's a bit hard.

@JoeZ. My program would work just as well if the system used ebcdic.  And as you know, a multitude of ebcdic systems use bash.  :-)

Maybe 34 bytes: :;y=$[x=!$?,x+x];echo $y$?$x$[y+y] Try it online!

Hmmm, Just 29 bytes: echo $[x=y=!x,y+=y]$?$x$[y+y] Try it online!

Clojure (177 49 characters)
EDIT
Just tripped over this question again and realized there's a much better way to approach this:
(Integer.(apply str(map count["AA""""A""AAAA"])))

which cuts the length down by quite a bit.
How it works:

(map count ["AA" "" "A" "AAAA"]) returns (2 0 1 4), which is a list containing the lengths of each of the strings in the argument vector.
(apply str (...)) converts the elements of the list (2 0 1 4) into the string "2014".
(Integer. (...)) converts the string from #2 ("2014") into the integer value 2014.
This snippet returns the value 2014.

Original
In the true Lisp-ish spirit that "too many parentheses are never enough" I present:
(Integer. (clojure.string/join [(+ (second (range))  (second (range))) (first (range)) (second (range)) (+ (second (range)) (second (range)) (second (range)) (second (range)))]))

How it works:
The function range produces a lazy sequence of numbers. If no starting point and ending point are specified the range starts at zero and extends infinitely in the positive direction; however, because it's a lazy sequence the numbers are not produced until needed. Thus, applying the first function to the result of the range function without arguments produces the value 0, which is the first element in the sequence 0 to positive infinity. Applying the function second to such a range produces the value 1. From there it's a simple matter of producing enough 1's and summing them up to get 2 and 4, then converting them (implicitly) into strings to join then together, then converting the resulting string back to an integer. (I find it amusing that this is actually longer than some of the Brainf*ck answers - and to add to the horror, it's also legible :-).
Share and enjoy.
:-)

I suppose that you don't need to convert back to integer, instead add an output function.

Do you need all that whitespace?

C# - 64 chars
class P{static void Main(){System.Console.Write('!'*'>'-' ');}}

pretty printed
class P
{
    static void Main()
    {
        System.Console.Write('!' * '>' - ' ');
    }
}

What if we use class P
{
static void Main()
{
System.Console.Write(2014);
}
}

@Amit, you can't use numbers in this challenge...

if an exit code is acceptable class a{static int Main(){return '!'*'>'-' ';}} is just 47 chars.

I like this one, my countdown numbers skills sucked so I ended up with ('d' + 'e').ToString() + Math.Ceiling(Math.PI);

VB.NET, 59 bytes
MsgBox(((Asc(vbTab) + Asc(vbTab)) & Asc("~")) / Asc(vbTab))

takes the ascii values of a Tab twice (18) concats the ascii value of "~" (126), giving "18126" and then divides the lot by ascii of a Tab (9) = 2014
Alternatively, you can do
MsgBox Asc("j") * vbKeyPause

i.e. ascii of "j" (106) * value of the constant vbKeyPause (19), for a total of 28 characters (less than half the original).

The exact same code works for VB6 too.

user14566 suggested this edit: `27 bytes: MsgBox(Asc("") & Asc(""))
=20 =14`

You can run this in the immediate window of VBA as ?Asc("j")*vbKeyPause, which shortens it up a bit.

newLISP - 61 characters
I can't seem to play golf in Lisp -- is it just too wordy?
(int(apply string(map(fn(c)(find c'(b c a e d)))'(a b c d))))

C, 31 bytes -- without a multi-character literal
main(){printf("%o",'\xe'*'J');}

Save 3 bytes by shortening main to f, since we don't require main to be used.

Clojure, 9 bytes
Inspired by the Matlab answer, converts char \u075e to an int:
(int \ߞ)

Haskell, 69 bytes
How do you get a number without using any digits in the source? Lots of people had already done it with characters or strings, so I decided to use pi, predefined in most languages. From pi, you can get the numbers 3 and 4 easily using the ceiling and floor functions. Then you can use some combination of addition, subtraction, multiplication, and maybe division to get 2014. Just by experimenting around, it would be easy to figure out a function that takes in 3 and 4 and returns 2014 (such as 4^4 * 4 + 3^3*3^3 + 4*(3^4) - 4*4*4 + 4 - 3 = 2014). This one's 70 characters:
main=print$(\x y->y^y*y+x^x*x^x+y*x^y-y*y*y+y-x)(floor pi)$ceiling pi

Now, that's fine, but writing a function like that isn't much different than just repeatedly writing floor(pi) and ceiling(pi). Is it doable with only one pi? Well, in Haskell, functions can be treated as Monads with an instance defined in Control.Monad.Instances:
instance Monad ((->) r) where
        return = const
        f >>= k = \ r -> k (f r) r

So you can use the bind function to pass one value into two different functions: g (f x) (h x) can be rewritten f >>= flip (g.h). id >>= f can be used to pass the one value twice into the same function: id >>= (^) for example is a function that returns x to the x power. The resulting program at 207 characters is more obfuscated than golfed, but it was fun to write:
import Control.Monad.Instances
main=print.((id>>=(^)>>=flip((+).(id>>=(+)>>=flip((+).(id>>=(-)>>=flip((+).(id>>=div))))))).floor>>=flip((-).(id>>=(^)>>=flip((*).(round.sqrt.fromInteger>>=(*)))).ceiling))$pi

Sclipting, 3 characters (6 bytes)
This outputs the string "2014".
꼣갱꽀

That's not 3 bytes though.

I didn’t say it was.

Well I can see that obviously, but the question actually asks for a byte count.

Byte count stands at 12, I believe.

No, it’s 6 bytes.

Pure bash 18
Without fork!
echo $[$[$#xd]#bbc]
2014

How about: printf %x \'— Just 13 characters (15 bytes).

@Isaac Awesome! I learn today this printf %d \'A newer seen before!

Pyth, 4 bytes
C"ߞ

Pretty straightforwards, just convert that character to an integer and print.

While this answer is now technically the shortest, I can't accept it because it was written in a language that didn't exist at the time of writing the question.

@JoeZ. I understand. No worries, the rules make sense.

J (13)
#.a.i.'_!!! '

Interprets the ASCII value of  _!!!  (95 33 33 33 32) as a binary number (it's weird that this is possible, I agree). This produces 2014.
J (15)
This one doesn't use any character strings. It's based on the weird coincidence that the sum of the first 46 primes is 4028: double 2014.
-:+/p:i.<:+:_bn

If anyone knows of a shorter way than <:+:_bn to represent 45 (preferably without strings), please let me know.

Racket, 18 bytes
(~a(+ #xa #xa)#xe)

Excel VBA, 51 bytes
MsgBox Len("aa") & Len("") & Len("a") & Len("four")

As it's 2015 you could add an extra character onto "four" for an extra byte.

Quetzalcoatl, 11 4 5 bytes
::ord('ߞ')

The box should be replaced by Unicode character 2014.
Edit
This is for an old version of Quetzalcoatl.
New version:
'ߞ'O

There is no ASCII character 2014... it has to be Unicode to go that high.

I'm not aware of any encoding in which ߞ would be a single byte.

Java 8, 33 bytes
()->Integer.parseInt("bbc",'\r');

EXCEL: 148 bytes
=POWER(ROW()+ROW(),(ROW()+ROW()+ROW())*(ROW()+ROW()+ROW())+ROW()+ROW())-(POWER(ROW()+ROW(),ROW()+ROW()+ROW())*(ROW()+ROW()+ROW()+ROW())+ROW())-ROW()
only works in A1.

It works anywhere in row 1.
Use the ^ operator instead.
You can also save a few using row 2 instead and a little algebra: =ROW()^((ROW()+ROW()/ROW())^ROW()+ROW())-ROW()-ROW()^(ROW()+ROW()+ROW()/ROW(. But then again, you can also save some by going all the way to Row 2014 too.

Of course, concatenation is also an option.

Sinclair ZX81 15 bytes 10 bytes
 PRINT CODE "=";CODE ":"

As the ZX81 has a non-ASCII compatible character set, the character code for = is 20 and for :  it is 14 - simples.

You've used a 1 in your answer.

There is no way around making a ZX81 BASIC program without using line numbers. You may take out the line number and run it in direct mode if you wish.

This is perhaps another reason why retrocomputing @ Stack Exchange should allow one-liners and code-golf therem, but apparently this would not constitute retro computing. Or something.

Correction on my last comment: there not therem.

R, 39 31 bytes:
x=T+T;x^(x*x*x+x)*x-x^(x*x)*x-x

R, also 39 31 bytes:
x=T+T;z=x*x;x^(z*x+x)*x-x^z*x-x

Thanks Scrooble!
More entertaining version: 46 bytes
z=pi;x=z*z;y=exp;j=z/y(z);floor(y(x)/(x-j-j))

Not especially efficient, but I had a lot of fun messing around with this.  I'm sure there's a shorter way using just those two numbers
Long-form, subbing in the variables:
floor(exp(pi*pi)/((pi*pi) - pi/exp(pi) - pi/exp(pi))
In real-person numbers:
floor(19333.69 / (9.869604 - 0.1357605 - 0.1357605)) =
floor(2014.328)

Save 8 bytes by getting 2 a shorter way.

Save a few in the entertaining one by eliminating some assignment.

And some more

Julia 0.6, 9 bytes
Int('ߞ')

Try it online!
Just for completeness' sake. Here's 2018 (same trick, different character):
Int('ߢ')

Try it online!
And just for fun, here's a function using bit shifting and arithmetic instead of using character codepoints (depends on this being Julia version 0.6, which seems an appropriately golf-y hack):
Julia 0.6, 47 bytes
(l=VERSION.minor,o=true)->o<<(l+l-o)-o<<~-l-o-o

Try it online!
Here, o=true evaluates as 1 during arithmetic. VERSION is an inbuilt constant containing the current Julia version, and VERSION.minor is 6 in this case. We left shift 1 by 6+6-1=11, giving 2048, then subtract 1<<(6-1)=32 and 1 and 1 from it, to give 2014.
2018 version would be:
(l=VERSION.minor,o=true)->o<<(l+l-o)-o<<~-l+o+o

-1 byte : '߾'-' '

MathGolf, 2 bytes
ID

Try it online!
Explanation
I   Pushes 20
 D  Pushes 14

The stack is printed in full on termination.

Keg, 2 bytes
ߞ

Keg auto pushes any characters that aren't instructions to the stack, and ߞ has a unicode value of 2014, which then gets printed.
Try it Online!

Raku (previously Perl6)
Raku interpret Unicode numbers as usual numbers, so:
In REPL – 7 bytes (3 chars):
⑳~⑭

Without REPL – 11 bytes (7 chars):
say ⑳~⑭

you can run it as rakudo -e "say ⑳~⑭"

Hexagony, 3 bytes
ߞ!

Try it online!
Or if you prefer code that terminates, here's 4 bytes:
ߞ!@

APOL, 38 bytes
-(⒏ l(*("eeeeeeeeeeeeeeeee" l("ee"))))
⒏ is a built-in constant that equals 2048, which is only 34 off from 2014. I use the length function to take the length of a 17-character string multiplied by 2 (which is shorter than a 34-character string) and implicitly print the result.

Fig, 4 bytes (UTF-8)
C/ߞ

Try it online!
Char casting, as most others.

Python 3, 27 bytes
I went through the effort of writing this script:
def to_base(n: int, b: int) -> str | None:
    if not 1 < b < 37:
        raise "Impossible base, try using a base between 1 and 37 (beginning and end not included)"

    if n == 0:
        return '0'

    possible_digs = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    res = []

    while n > 0:
        res += possible_digs[n % b]
        n //= b

    return ''.join(reversed(res))

def to_base_letters_only(n: int) -> str:
    bases_found = []
    for b in range(11, 37):
        c_n = to_base(n, b)
        digs_l_10 = False

        for d in c_n:
            if d in "0123456789":
                digs_l_10 = True
                break

        if not digs_l_10:
            bases_found.append([c_n, b])

    print(bases_found)

    if bases_found == []:
        return "No conversion found"
    return f"Conversion found: {(min_base := min(bases_found, key=lambda c: len(c[0])))[0]} in base {min_base[1]}"

print(to_base_letters_only(2014))

Which found BBC in base 13, which is 2014, without using arabic numerals.
So, my answer is this:
print(int("BBC",ord("\r")))

Try it online!
\r is Unicode 13.
Fun fact: This answer is 1 byte behind the Python 2 answer, because of print needing parentheses.
That script can also be used to find solutions for other numbers.

Normally we put the code first and the explanation afterwards

You could save a byte by switching to Python 2: print int("BBC",ord("\r"))

I already pointed out that.

Fortran: (43 27)
print*,z'FBC'/len('hi');end

Thanks to Hristo Iliev, the above is about 40% smaller! z'FBC' returns the decimal form of that hex value (which is 4028), len returns the length of hi (i.e.,2).

Original answer:
print*,ichar(',')*ichar(',')+ichar('N');end

Converts the string , and N to ASCII values: 44 & 78 respectively: 44**2 + 78 = 1936 + 78 = 2014.

Shorter version using hexadecimal literals: print*,z'FBC'/len('hi');end.

@HristoIliev: Totally forgot about printing hex via z! Thanks a bunch!

PHP (21 chars)
<?=ord('').ord(''); //These are not empty strings ;)

If you don't believe it, see the proof.

That looks like 19 characters to me.

(Oh wait, nonprintables.)

If it contains non-printables, you should provide a hex dump or list them.

 Python 51
Using true = 1 and false = 0
t=True
print str(t+t)+str(t-t)+str(+t)+str(t+t+t+t)

clever. 40 chars in PHP: $t=true;echo $t+$t.$t-$t.$t.$t+$t+$t+$t;

Damn just wrote that while reading the answers well done,

Python, 30 chars
s=int('RZ',ord('$'));print s+s

2014 => 2 * 1007 => RZ in base 36 => ascii code for $ character
In interpreted mode, without the print statement it is 24 chars:
s=int('RZ',ord('$'));s+s

Julia, 13 characters
('x'-'e')*'j'

In Julia, most arithmetic operations, when applied to a single character, convert this character to its ASCII integer value. x, e and j are respectively 120, 101 and 106, therefore (120-101)*106  is 19*106=2014.
julia> ('x'-'e')*'j'
2014

Edit: 11 characters, thanks to Glen O
A different choice of characters allows us to skip parentheses:
'.'*'.'-'f'

Just thought I'd point out that a different sequence can save you a few characters. For instance, '.'*'.'-'f' is only 11 characters.

@GlenO thanks! I added it as an edit.

Bash, 29 bytes
Bash without using external programs:
echo $((x=++y+y))$?$y$((x+x))

Reduce to 25 bytes by using: echo $[y=++x+x]$?$x$[y+y].

~-~! (No Comment), 41
Pretty basic solution.
'=~~~~~:''=~~,','@'':@''-~~:@''-~:@''+~~:

Pretty good for just 8 unique characters, eh? xD So this could theoretically be stored in 123 bits, or ~15.4 bytes.

k [16 chars]
(*/"i"$".,")-@""
2014

Explanation
Get the ASCII value of ",.".
"i"$".,"
46 44

Find the product
*/"i"$".,"
2024

Get the data type of char.
@""
10h

On running the complete code (2024-10)
(*/"i"$".,")-@""
2014

12 chars: +/&" ~~~~h'";
6 chars, 7 bytes, unicodey: \i$"ߞ"`

><> (9 bytes ASCII)
In pure ASCII,
'd!:'*+n;

This pushes d, !, and : to the stack, then multiplies the numerical values of top two entries, and adds the value of the last entry before outputting the value on top of the stack as a number and ending.
Using Unicode this can be reduced to 6 bytes:
'ߞ'n;

Simply outputs the numerical value of ߞ and ends.

You could shorten 'ߞ'n; to 'n;ߞ, I believe.

C#, 4 characters, 5 bytes
+'ߞ'

Note: you need LINQPad to run it, not Visual Studio. LinqPad is good for CodeGolfing in C#.

It's 4 characters, yes, but 5 bytes.

@JoeZ. ok, updated to reflect the number of bytes. Still way better than previous 63 and 64 bytes solutions.

Insomnia, 7
Each line is one program doing the same thing: print 2014 to output stream.
e}u#Hi-
e}u#Hs-
e}u#H}-
e}g#*i-
e}g#*s-
e}g#*}-
e}gKHi-
e}gKH}-
e}gKxi-
e}gKxs-
e}gKx}-
e}u#dK-
e}u#eK-
e}u#fK-
e}gKdK-
e}gKeK-
e}gKfK-

JavaScript, 24 bytes
A bit long, but no idea how this way got left out...

alert("ߞ".charCodeAt())

Explanation
The character ߞ is obtained by doing String.fromCharCode(2014) . Thus the code is actually just converting that character back to its character code and alerting it.
Thanks to hsl for this shorter version

That code doesn't work. Did you mean alert("ߞ".charCodeAt())?

@hsl String.charCodeAt is present only in Firefox, it seems. But I'll use charCodeAt since its multi browser and shorter . Thanks!

Python 2 (19 bytes, ASCII only, CPython-specific)
print hash("w_'qe")

Tested only on 64-bit, but I assume/hope that since 2014 is small and positive the results would be the same on 32-bit? Originally tested on Python 3, but ProgramFOX confirms it also works on Python 2.
Python 3 (31 bytes, ASCII only)
print(ord("\N{NKO LETTER KA}"))

Quite fond of this one, even though better solutions exist. The equivalent Python 2 code is no shorter, as it required a u string prefix.

I tested on Python 2.7, and it works fine there; so you can save one character.

I found the same python 3 version, but shorter (16 bytes) as I didn’t restrict myself to ASCII :print(ord('ߞ'))

CMD - 42 bytes
set/aa=f
set/a%a%xAAA-%a%xFF-%a%xFF-%a%xCE

The 'trick' is that when using the /a switch on the set command, letters (and other invalid characters) are evaluated as 0. I then just use hexadecimal to evaluate 2014. The 0 is needed because in CMD hexadecimal must be expressed with the leading 0x. There is almost definitely a shorter way to get to 2014...
%a%xAAA - %a%xFF - %a%xFF - %a%xCE = 2730 - 255 - 255 - 206 = 2014

Python (30 chars)
(10 + 9) * 106 = 2014
(ord('\t')+ord('\n'))*ord('j')

Hassium, 77 Bytes
Really excited about this one. It gets Math.pi and divides it by itself in variable a (1), then uses increment and basic math operators to get it to 2014.
use Math;func main(){a=Math.pi;a/=a;print(++a)print(a-a)print(a/a)print(a*a)}

Run online and see expanded here

T-SQL 27 bytes
PRINT ASCII('')*ASCII('j')

Note that the character that isn't rendered here is the DC3 (CHAR(19)) in the first set of quote marks.  It's unicode U+009F which, it would appear, doesn't copy and paste here too well but I can assure you it works in SQL Management Studio.

Jolf, 3 bytes
(It's almost 2016. Language obviously postdates question.)
@ߞ
@  Get charcode of next character
 ߞ charcode 2014

JavaScript, 9 bytes
btoa`ÛMx`

This Base-64 encodes ÛMx to make 2014.

Lua, 32 bytes
print(#'XX'..#''..#'X'..#'XXXX')

This makes use of the length operator # used on string literals to get numbers, and the lengths are concatenated (with ..) to print 2014

Fourier, 12 bytes
Non-competing: Fourier is newer than the challenge
I know I'm two years too late, but it seemed like a fun challenge to do in Fourier.
^^ovvo^o^^^o

The command ^ increments the value of the accumulator (which starts at zero) and v decrements the value of the accumulator.
Try it online!

Straw, 13 bytes
(…………………σ)«$>

« sum the codepoint of all characters in a string, $ convert from unary to decimal and > is the print operator.
Try it online

Jelly, 3 bytes
⁽¥Æ

Try it online!

@OliverNi https://codegolf.meta.stackexchange.com/questions/12877/lets-allow-newer-languages-versions-for-older-challenges

Brain-Flak, 50 bytes
((((((()()()()()){}){})){}{}()){()()({}[()])}{}())

Try it online!
Brain-flak is great for restricted source challenges because there are only 8 valid character to begin with: brackets. (e.g. []{}()<>).
This was made possible with the help of @ASCII-only's integer metagolfer, which is currently hosted online at brain-flak.github.io/integer

Using the integer metagolfer in WheatWizard's brain-flak optimizer I found ((((((()()()()()){}){})){}{}()){()()({}[()])}{}()) which is 4 bytes shorter.

The challenge requires languages in which 0123456789 are valid tokens. Does Brain-Flak satisfy this requirement?

Brain-Flak, 46 + 3 = 49
Try it online
(([()]([(()()())]((((({}){}){}()){}){})))()())

This one uses the -A flag for +3 bytes.  It pushes the ASCII values for 2014 to the stack which outputs as 2014 in ASCII mode.

The challenge requires languages in which 0123456789 are valid tokens. Does Brain-Flak satisfy this requirement?

@Dennis yes @lt flags use decimal literals

Lean Mean Bean Machine, 53 bytes
\\\
O))\ \
#)) o )
 ))u ))
 ))  ))
 ))  )u
 ))  ~
~~~

I like this.

J, 21 bytes
,":,.$,:~}.,:,:'golf'

Try it online!
               'golf'  One dimensional array
           ,:,:        Itemize twice (1x1x4 array)
         }.            Drop first element (0x1x4)
      ,:~              Append to itself as distinct items (2x0x1x4)
     $                 Get dimensions (2 0 1 4)
   ,.                  Flatten items, essentially prints 2014 vertically.
                       (so there are no spaces)
,":                    To strings, flatten.

20 bytes
#.(#_),,~(,~,~#_),%_
#.(#_),,~(,~$,._),%_

15 bytes
do'bbbc',~":_bd

11 bytes
,":_bk,:_be

><>, 7 bytes
aa+nen;

Try it online!
aa                      +                           n                                    e                 n                                    ;
^^                      ^                           ^                                    ^                 ^                                    ^
Push 10 to stack twice, add top two items in stack, print top item in stack as a number, push 14 to stack, print top item in stack as a number, stop.

6 bytes: eaa+n<

JavaScript, various solutions: 131, 44, 43, 30, 28, and 16 characters (16 bytes)
Original answer:
JS-fuck inspired JavaScript, 131 characters:
(!![]<<(!![]<<!![])<<!![]<<(!![]<<!![])<<!![]<<(!![]<<!![])<<!![]<<(!![]<<!![]))-((!![]<<!![])<<!![]<<(!![]<<!![])<<!![])-!![]-!![]

Shifty truth edition:
(true<<(true<<true)<<true<<(true<<true)<<true<<(true<<true)<<true<<(true<<true))-((true<<true)<<true<<(true<<true)<<true)-true-true

Addenda
Edit: I'm having too much fun with this... combining type coercion with hex strings, 44 characters.
a=+[];(a+'xade')-(a+'x'+(+!a+!a+!a)+a+a)
a=+[];(a+'xade')-(a+'x'+(++a+a+a)+--a+a)

a is 0, 0xade is 2782. We need to subtract 768 to get 2014. 768 is 0x300. (a+'x'+(+!a+!a+!a)+a+a) and (a+'x'+(++a+a+a)+--a+a) are two ways of producing "0x300", so the final result is "0xade" - "0x300", which JavaScript coerces back to numbers, resulting in 2014.
Here is a radix 36-based parseInt solution, 43 characters:
(a=+[]);parseInt(++a+'jy',a+++a+++''+(a+a))

First, we initiate a variable a at 0 (clearly this requires non-strict mode), increment it to 1, concatenate with 'jy' - 1jy happens to be 2014 in radix 36. The quickest way call parseInt at this radix is to generate the string "36" and abuse type coercion again: a++ + a++ results in 3, with a set to 3 as well, which means (a+a) is 6, so a+++a+++''+(a+a) results in "36", resulting in 2014.
After coming up with these two solutions I started looking at other JavaScript answers (thanks for the tip, Scrooble!), to see if combining ideas from other people gives interesting results.
Zaq's approach can be shortened with the hex string trick to 28 characters:
-~[]+!![]+[+[]]+ +(+[]+'xe')

How it works: -~[]+!![]+[+[]] results in "20", +[]+'xe' results in "0xe". +"0xe" is 14, so "20" + +"0xe" becomes "20" + 14, which becomes "2014".
Now here's a fun fact: 20 in hexadecimal is 0x14! Oh, and 14 is 0xe. So +('0x' + 0xe) produces 20. Combining this with the above results in a 30 character solution:
_=+[]+'x';+(_+ +(_+='e'))+[+_]"

Finally, initially inspired by Dan Prince's answer I came up with what might be the shortest possible JavaScript solution abusing hex-strings, at 16 characters:
+[]+'xfbc'>>!![]

0xfbc is 4028. Shift right by one, and we have 2014. If anyone knows of a shorter way to generate 0 or 1, let me know.
Edit2: having exhausted the hexadecimals, we should of course also look at binary and octal notation!
// +'0b11111011110'
+(a=+[],a+++'b'+(a+(((a+=[a]+a+a+(+[])))+a)))
// +'0o3736'
+(a=-~[],a=a+++a+++''+(a+a),+[]+'o'+((+a+1)+a))

Welcome to the site, and nice first post!

Thanks. Do you know of any ways to show all answers on one page so I can quickly check for duplicates next time? ;)

@Job the stack snippet in the question body lists by shortest solutions and the shortest solutions by language. I've never been able to see them on mobile/in the app, though, so your mileage may vary.

@Job You can do a search with inquestion. For just about all questions, there'll be just one page of results. The 17005 is this question's ID, found in the URL.

Thanks for the tips! Added some more solutions :)

Pyt, 27 bytes
ɳąḞḞ⬠⬠⬠π⎶⁻⦋ĐąžΠ²+ĐŚřƩ½*-⁻⁻Ɩ

Try it online!
Not exactly a serious contender, just had some fun.

ɳ             push '0123456789' as string
 ą             convert to array of digits [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
  ḞḞ            for each item in array, replace with xth fibonacci # (2x) [1, 1, 2, 3, 8, 34, 377, 17711, 9227465, 225851433717]
   ⬠⬠⬠          for each item in array, replace with xth pentagonal # (3x) [1, 1, 1820, 66045, 240027425, 29321506727800, 6947548864499411875070L, 165405818231059923692911546880492501L, 898044801648686628863443901192030771814779461710865094720L, 115670237695821250427139838385782853032222541808893547195455834936957002151009052998969975100L]
       π⎶⁻           push pi, round, and decrement (2)
          ⦋         get the 2th element of that list (1820)
           Đ         Duplicate 1820
            ąž        make an array of digits and remove zeroes [1, 8, 2]
              Π        multiply together (16)
               ²+       square and add (2076)
                 ĐŚ      Duplicate and sum digits (15)
                   řƩ     make a range from 1 to 15 and sum all (120)
                     ½*    multiply by 1/2 (60)
                       -    subtract  (2016.0)
                        ⁻⁻   decrement (2x)   (2014.0)
                          Ɩ   cast to integer (2014)
implicit output

JavaScript (ES6), 52 46 bytes
t=-~'';console.log(`${++t}${t-t}${t/t}${t*t}`)

Try it online!
Will log through console.log the number 2014 as a string.
Thanks to Jacob for multiple optimizations saving 6 bytes
Explanation
t=-~'';

In JavaScript, ~~ will convert the proceeding value to a number, in this case, true equates to 1.
Set the value of t to 1 by using ~ on an empty string, which would equate to -1, then take the opposite of that number, 1.
For more info about tilde in JavaScript, see this article.
console.log(` ... `)

Logs the template string ... with ${} expressions available, where ... includes:
${++t}

Sets w to t+t, which would be 2, which would return the number 2. Added to string.
Set t to itself + 1, and display the final result, 2
${t-t}

Displays t-t, which would be 0, which would return the number 0.
${t/t}

Takes the value of t and divides by itself, returning 1.
${t*t}

Takes the value of w*w*w*w-w (or w ^ 4 - 1), where w (as previously set) is 2, and subtracts w from it, and returns the result. Added to string.
Takes the value of t*t (or t ^ 2), where t (as previously set) is 2.
The added expressions equate to 2014, which ... is in the log.

Technically you don't need the ~~ before the true

t=-~'';console.log(\${++t}${t-t}${t/t}${t*t}`)`

Lost, 57 23 22 bytes
<^/*(
 )/@+
">>v
^?%<<

My first Lost answer. Thought I'd start with an easy one.
Byte-count more than halved (-35 bytes) thanks to @JoKing.
Try it online or verify that it's deterministic.
General explanation about Lost:
Let me start with an explanation of Lost itself. Lost is a 2D path-walking language. Most 2D path-walking languages start at the top-left position and travel towards the right by default. Lost is unique however, in that both the start position AND starting direction it travels in is completely random. So making the program deterministic, meaning it will have the same output regardless of where it starts or travels, can be quite tricky.
A Lost program of 2 rows and 5 characters per row can have 40 possible program flows. It can start on any one of the 10 characters in the program, and it can start traveling up, down, left, or right.
In Lost you therefore want to lead everything to a starting position so it follows the designed path you want it to. In addition, you'll usually have to clean up the stack when it starts somewhere in the middle.
Program explanation:
The 22-bytes program is similar as the previous 23-bytes program below, but with a smarter path to save that byte:
v<<<<<>>>>>
>%?"^ <"*+@

Let me start with an explanation of the 23-bytes program:
The "^ <" will push the character-codepoints for the three characters in the string, being 94 32 60 respectively. The * multiplies the top two, and + adds the top two of the stack, so it becomes 94+(32*60), which results in 2014.
The @ will terminate the program, but only if the safety is 'off'. When the program starts the safety is always 'on', otherwise the program starting at the exit character immediately terminates without doing anything.
The % will turn the safety 'off'. So as soon as the % is encountered and the safety is 'off', the program can be terminated with an @.
The ? is to clean up the stack if it started somewhere in the middle.
And finally the v<<<<<>>>>>, > and use of ^ < in the string are to lead the program path towards the correct starting position for it to correctly print 2014. Note that the top line could have been v<<<<<<<<<<, but that the reversed part >>>>> will wrap-around to the other side, making the path shorter and therefore the performance slightly better. The byte-count remains the same anyway, so why not.

Now for the 22-bytes solution, and how it actually is the same as the 23-bytes solution, but with a different path.
The arrows are still used to lead the path into the given direction. The / are used as a mirror. So if we go from right to left and encounter the /, it will continue downwards; if we go from the top to the bottom and encounter the /, it will continue towards the left; etc.
The ( will pop the top value on the stack and push to to the scope, and the ) will do the reversed: it pops from the scope, and pushes it back to the stack.
So regardless of where we start and in which direction we travel, the path leads towards the first < of the bottom row. From there, the program flow travels in this order:
%?^        Direction changed upwards
" <^" <    Direction changed towards the left
(*/        Direction changed downwards
/          Direction changed towards the left
) +@

So it will:

Turn the safety 'off' with %;
Clean the stack with ?;
Push the character-codepoints for " <^", which are 32 60 94 respectively;
Pop the 94 and store it in the scope with (;
Multiply the 32 60 with *, resulting in 1920;
Push the 94 from the scope back onto the stack with );
Add the 1920 94 together with +, resulting in 2014;
And then terminates the program with @, implicitly outputting the top of the stack.

Why not calculate the 2014 code point and skip the -A flag? 23 bytes

@JoKing Ah, didn't thought about that.. Thanks a lot for halving the byte-count! :)

22 bytes. How it actually works is a bit weird...

@JoKing Oh, smart! That 23-byter was pretty obvious and I can't believe I missed it now. But that 22-byter is very smart and not something I would have thought of myself. Well done! It's only my first Lost answer, so hopefully I will get better at it. Btw, out of curiosity, is there a reason the language is lacking a divide and modulo operator?

shrugs Minimalism?

bc, 7 bytes. Try it Online!!
K*ZZ+Y
bc,  8 bytes. Try it Online!!
K*A*A+E
Which needs 14 bytes to run in bash:
bc<<<"K*A*A+E"
In bc the upper (single) letters maintan their meaning as a number in 10-36 range in any input base.
A previous approach changed the input base:
echo ibase=D\;BBC|bc
Make numeric base 13 (D) and print BBC in that base --> 2014.

Python 3, 16 bytes (15 characters)
print(ord('ߞ'))

Try it online!
Explanation
ord returns the decimal Unicode of a character, and the decimal Unicode of ߞ happens to be 2014.
Python 2?
I tried doing the same thing in Python 2 (print ord('ߞ')), which would be 1 byte less, but this doesn't work. Why? Well, in Python 3, len('ߞ') returns 1, so everything is fine. However, in Python 2, it returns 2. And since ord only takes a string of length 1, Python 2 doesn't really like that: TypeError: ord() expected a character, but string of length 2 found

C#, 25 bytes (24 characters)
Console.Write((int)'ߞ');

Try it online!
Explanation
The decimal Unicode of ߞ is 2014, so you can just cast it to an int and 2014 is printed.

Welcome! Though admittedly, the answer is simple, in general it is best to add an explanation or link to an online interpreter. Code-only answers are usually automatically flagged as low-quality, meaning that those answers have to be reviewed by someone.

@mbomb007 yeah i was thinking about adding a link to tio.run, i'll do that

05AB1E, 22 10 8 7 6 bytes
T·žvÍJ

Try it online!
How it works:

T pushes 10, · doubles it, we get 20.
žv pushes 16, Í subtracts 2, we get 14.

Lastly, we concatenate them using J!

Hi, welcome to PPCG! See here for an existing 6- and 5-byte 05AB1E answer. But to golf your approach a bit more: 1. you can remove all « and only have a single J (join) at the end. 2. You could save values in between with © and use it later on with ® (i.e. žh¦¦©нžhнžh¦н®¦¦нJ). 3. There are builtins for the numbers 0, 1, and 2 if you don't change them later on, which are ¾, X, and Y respectively, so an alternative 6-byter could be Y¾XY·J (push 2; push 0; push 1; push 2 and double it; join all values on the stack together).

(this last one doesn't really look like your initial approach anymore, but just stating it as example). And if you haven't seen it yet, tips for golfing in 05AB1E might be interesting to read through. :)

Digits aren't allowed though. Otherwise just Ž7æ (compressed 2014) or 2014 itself would be enough. :)

@KevinCruijssen Oh, how silly. I forgot.

I've found a 10 byte solution now. I don't think there might be any room for improvement.

+1 from me for effort. :) PS: žz could be ₁ (256), ₂ (26), or ₅ (255).

Actually, I found a 8 byte solution. I changed 0 and 1 to 10 and reversed it.

Flobnar, 17 14 bytes
<+\@:!
+<>.!..

Try it online!
Explanation:
   @       Start, going left

  \        Push to the stack the value from the bottom row
  >

   .!..   Several print statements we will get back to
<   :!    Add the not of the top of the stack to itself
+         This is !0+!0 = 2

      .   Print the 2

     .    Print the result of the print (0)

   .!     Print the result of the not of the print (1)
  \       Continue forward after pushing the zero to the stack

<+   :!   Add the same 2 from the beginning to itself
+<        (!0 + !0)+(!0 + !0) = 4
          And print implicitly as the last value

Python, 52 49 chars
from math import*
print(int((e*pi+e)**pi+e/e+e))

Works in Python 2 and Python 3.

MAWP, 16 bytes
!!M:!!A:!:!!M!M:

chr:pos:stack
! : 1 : [1,1]
! : 2 : [1,1,1]
M : 3 : [1,2]
: : 4 : [1]
! : 5 : [1,1]
! : 6 : [1,1,1]
A : 7 : [1,0]
: : 8 : [1]
! : 9 : [1,1]
: : 10 : [1]
! : 11 : [1,1]
! : 12 : [1,1,1]
M : 13 : [1,2]
! : 14 : [1,2,2]
M : 15 : [1,4]
: : 16 : [1]

Try it!

Ayy, somebody actually uses my language :D

I tried using _ in the online interpreter, but for some reason it didn't work. I made a Gtihub issue on it. I'd like to solve some more puzzles, but that operator is causing a bit of a problem.

Ok, will check and fix. Thanks for notifying me!

Labyrinth, 12 bytes
))!!)!))))!@

Try it online!
))))_#!!#!!@

Try it online!
))_#"
 ) !!@

Try it online!
))!"
)
)#!!@

Try it online!
I tried both linear and complex layouts, but I can't figure out how to remove a single byte from any of these programs.

11 bytes by reusing the prints. And an alternative

Hexagony, 18 bytes
g{A*'-"'-'"Av<@!}/

   g { A
  * ' - "
 ' - ' " A
  v < @ !
   } / .

I had trouble getting it to fit in 19 bytes (side length 3) because I was always 1 short, then I rearranged my memory accessing to be 1 shorter, which also allowed me to use a very efficient layout. Then I was able to shift a no-op somewhere in the code to the very end saving a byte.
Try it online!

Husk, 4 bytes
c'ߞ

Try it online!

Python 3, 49 bytes
for i in['..','','.','....']:print(len(i),end='')

Python 3, 15 bytes
print(ord('ߞ'))

Welcome to the site, and creative first answer! You can improve your score by remove the spaces before the [, before the print and before the end: Try it online!

@cairdcoinheringaahing thanks for notice me!

Scratch, 82 bytes
when gf clicked
say(join(length of[The year two thousnd])(length of[and   fourteen

Happy 2021! Scratch solutions make me happy though so take your upvote :P

Out of almost 300 answers, I was surprised that nobody had done Scratch yet!

If functions are allowed, you can do 73 bytes: define say(join(length of[The year two thousnd])(length of[and   fourteen

With \n a newline

@UndoneStudios Scratch's syntax is the bane of my existence! However, I generally prefer full programs over functions, since they can be easily viewed in a project. Also, OP only specifies "programs." Thanks, though!

Like I said "If functions are allowed" so well, welcome!

CSASM v2.4.0.2, 83 bytes
func main:
push "  "
len
print
push ""
len
print
push " "
len
print
push "    "
len
print
ret
end

The only way to push numbers to the stack without using numbers is to create str instances and then get their lengths.

Alphabetti spaghetti, 15 bytes
aiioaoaioaiiiio

Try it online!

11 bytes with iioaoioiiio or iiuuiuiiiio.

not sure if others have used this one already :

from any UTF-8 aware shell

printf %d \'ߞ

13-chars spanning 14-bytes. the following are equivalent forms :
printf %d \'$'\xDF\x9E'
printf %d \'$'\737\636'

ACCUMULATOR, 2015 28 bytes
-1987 bytes thanks to a near-escape from the Orwellian thought police!
AAAAAAAAAAAAAAAAAAAACMMMMMMO

This is pretty simple:

20 As increment the accumulator: 20
C concatenates it to itself: 2020
6 Ms decrement: 2014
O outputs

Old version here

Desmos, 40 26 25 Bytes
b=\ln ee
k=bb
k^kkb-kkb-b

Desmos graph link
Now smaller by a whole 1 more byte! Yaaay...
Version that needs each line to be pasted individually, 24 23 bytes
b=lnee
k=bb
k^kkb-kkb-b

Could you provide a Desmos graph link in order to easily be able to verify your code? Also, you need to provide the byte count of the program in the header of the post.

You can shorten b=cos(\tau) to b=lne.

For reference, your solution is 40 bytes at the moment. I have taken pretty much the same calculations which you have in your code, but shortened it considerably to 24 bytes: 24 bytes, Try It On Desmos!

It looks like b=lnee doesn't work since there has to be a space between the function and the characters otherwise desmos does... weird stuff and you have to have a backslash before ln ee so that desmos doesn't just see it as 2 undeclared variables. (also, your last line uses the number 2 but that can be swapped out for b)
Still, thanks for the help.

If you paste in one line at a time instead of the whole thing at once (so paste in b=lnee in the first box, k=bbb in the next line, and kkkbb-kbb-b on the next line), then you will see that it works.

Ah, I see, thanks.

It is the de facto way of pasting in Desmos expressions due to the differences between single line pastes and multiline pastes. It turns out that single line pastes result in golfier code than multiline. What I'm saying is that you don't have to separate your answer into two different versions, one for multiline and one for single line pasting, because the single line pasting version is completely valid.

Java, 18 bytes
Just for the sake of using hashCode().
"bmgjagr".hashCode() = 2014
"?=".hashCode() = 2014
"zsjpzdq".hashCode() = 2024

In order to find a candidate string, I iterated over all the strings of 7 and less characters in the range ['a', 'z']. It represents a set of 8'353'082'583 strings and thus a good probability that one of them has the required hashCode.
Thanks to anatolyg, using more characters allows a much simpler string : "?=" (ASCII 63 and 61), 63 * 31 + 61 = 2014.
79 bytes full class:
class C{public static void main(String[]a){System.out.print("?=".hashCode());}}

18 bytes lambda:
x->"?=".hashCode()

I thought 2024 was the required output this year. Also I checked your link and I couldn't find how it helps me to shorten ".hashCode()".

The hashing algorithm is very simple: it's a series of additions and multiplications by 31. No need for brute-forcing. You can use, for example, 2014=63*31+61.

Ok, I got it now, I first wanted to use only a-z characters which make it much more complicated, but "?=" works as well.

Python 3, 54 bytes
n=mp.dps=ord('㤃');print(str(mp.pi)[n-len('year'):n])

Try it online!
How it works
Extracts the year \$2024\$ beginning at the \$14591^{st}\$ digit of \$\pi\$
3.14...27617285830243559830032042024512072872535581195840...
The next 10 years that eventually appear in Pi's decimal expansion!

https://en.m.wiktionary.org/wiki/%E3%A4%83

ANSI C - 95 47 52 characters

#include <stdio.h>
main() { printf("%i", (('a' + 'a')/'a') * ('\a' + '\f') * ('<' - '\a') ); }

This program uses characters to initialise integers and multiplies: 2 * 19 * 53.

    #include
    main(){printf("%i",'\aÞ');}
This program initialises an integer using charaterbytes and prints it. '\aÞ' is the bitpattern 00000111   11011110 this is also the bitpattern of 2014.

Disclaimer: this was made on a windows system with visual studio. This code depends on a lot of things, including - How your compiler endodes the characters you input. Þ has an ascii value of 222 (or its negative equivalent), this may vary depending on your system. The notation int a = 'abcd'; is in itself evil and depends on how memory is handled on your system - this includes endian issues. int a = '\0A'; a is 65 on my system but may be 16640 on your system.

main(){printf("%i",('C'-'A')*('T'-'A')*('v'-'A'));}

I went back to Version one and multiplied 2 * 19 * 53. This version uses only one byte at a time so it is endian compatible. Also it uses only characters in the range of [0 - 127] to be compatible to all systems.

You don't actually need to include stdio.h, most compilers will give a warning but include it for you.

This code yields 7 for me with tcc and 508830 with gcc. clang gives an error: character too large for enclosing character literal type.

@nyuszika7h ... 7 is equivalent with \a - this could be a problem with passing to the printf function, it would be interesting for me to see if int a = '\aÞ' is 2014 using tcc. 508830 looks very strange to me - i would guess some endian thing but n * 256 + 7 can never be that number. So this illuminates how string this code depends on the system.

Racket: 20 (19 chars)
(char->integer #\ߞ)

ߞ is a unicode character that has 2014 as it's code.
This abuses the fact that every top level form gets its evaluation printed to stdout. This is quite unique amongst LISPs which usualy only have this behaviour in the REPL and not when running programs.
Scheme: 29 bytes (28 chars)
(display(char->integer #\ߞ))

So 29 bytes, then?

@JoeZ. Yes. I was just about to update it :)

At least in Racket, you don’t need to call display — the REPL will print the result of char->integer automatically — which would save you another 9 chars.

@MatthewButterick Unlike Scheme and CL #!racket actually display every top level result to stdout even when running a compiled program. It's quite annoying but I can abuse that. thanks :)

Still beat it, I’m afraid

Solution 1
Octave/Matlab (55 chars)
a=pi;b=a*a;disp(ceil(a^a^a/a/a/a-b*b*a-a^a*b+b*b-b-b));

Solution 2
PHP (9 chars without tags, 12 with them Actually 2022 because of the new lines involved)

<?=__LINE__;

That would be 2013 newlines followed by <?=__LINE__ for 2024 chars, not 9.

@PeterTaylor I didn't write 9 chars to win the challenge (since there are lots of shortest answers!), but because the actual code is that <?=__LINE__;?>, which I thought would be funny :) Nvm, I'll edit that.

If it doesn't work without the newlines, they're part of the "actual code".

I guess you're right, edited.

In PHP, you can skip ?> at end of the program. But interesting idea with __LINE__, even if it's ridiculous for such huge number.

 C/C++ 39
main(){printf("%d%d",':'-'&',':'-',');}

ASCII for: ':' = 58, '&' = 38, ',' = 44.
Using that, 58-38 = 20 and 58-44 = 14.

120 characters in Squeak Smalltalk trunk (4.5).
I did not search the shortest, but kind of graphical solution:
((Text string:'Happy\New year'withCRs attribute:TextEmphasis narrow)asMorph borderWidth:Float one+Float one)bounds area

It depends on font, margins, and so is quite fragile, but at least for me it worked.
In Squeak 4.4, it works with lowercase 'happy\new year'.

Game Maker Language, 22
show_message(ord("ߞ"))

C - 44 characters (85 with headers)
What, no one is abusing strings yet?
#include<stdio.h>
#include<netinet/in.h>
main(){printf("%u",ntohs(*(int*)"\a\xde"));}

Interestingly, this is a special case where neither character is printable, but their special code doesn't involve a number.
If we want no warnings, it needs to become 55 (96) characters:
#include<stdio.h>
#include<netinet/in.h>
int main(){return!printf("%u",ntohs(*(int*)"\a\xde"));}

JavaScript, 49 Chars
A mathematical JavaScript version making use of only PI and E as source numbers.
(m=Math).pow(e=m.E,e*(p=~~m.PI))/m.sqrt(p)+e+e|""

... mmmm PIE.
Oh and just in case implicit returns are vetoed (56 Chars with alert):
alert((m=Math).pow(e=m.E,e*(p=~~m.PI))/m.sqrt(p)+e+e|"")

C++ - 63 bytes
I'm not sure if this method has been used, but I designed this myself anyway:
#include<iostream>
int main(){std::cout<<int('&'*(','+'\t'));}

You can use int instead of toascii for all of those to save a lot of characters.

@JoeZ. OK, I didn't think of that.  How can I calculate the bytes?

Save your program as a text file, and then view how many bytes it is in the file manager.

2 is not allowed.

Oops, I'll fix that.

@DankoDurbić Fixed.

I only count 62 bytes (using http://www.charactercountonline.com/). Also, can \t be replaced with the literal tab character?

@Quincunx Literal tab character?

@hosch250 is that not what \t stands for? (tab. #9 in ascii: http://www.asciitable.com/)

@Quincunx Yes, I can use plain 9.  However, I can have no numbers in my source code, so I cannot use 9 instead.  TAB does not work.

Clojure - 22
(apply *(map int"j#"))

(note: the # is ASCII character 19, Stack Overflow doesn't seem to like this but it's valid Clojure source...)
Clojure - 36
(dec(reduce +(nnext(range(int\@)))))

vba (immediate window), 38 26 13
using regular ascii characters (no funny typing needed)
?&ha+&ha&&&he

26
?val("&hfbc")/-(true+true)

38
?year((cdbl(asc("ê"))*cdbl(asc("²"))))

find a date that can be represented as a number, and select the year from that (in this case, Jan, 13, 2014)

have to use cdbl, as it assumes signed int, and overflows

How do you write that in the Immediate window? (I assume that is what you mean by “direct window”.) If I copy-paste it I get “?year((cdbl(asc("e^"))*cdbl(asc("^(2)"))))”. (Copy-pasting “ê” and “²” from charmap.exe results “?” both.) And of course, that way the calculation not gives 2014.

I used ?chr(234),chr(178) to get the characters, or you can hold down the ALT key and type 234 (and 178) and release the ALT to get each character

With Alt+234 I get “r”, with Alt+178 I get “¦”. Of course, it works with the chr() function. Anyway, nice trick to use year() this way.

ok.. I was thinking back to DOS days - now it's 0234 and 0178, but I found another shorter way now

Thanks, it works this way. Although here appears “ȩ” and “¸”, the calculation is correct.

Clojure, no unicode tricks (49 characters/bytes)
Uses the fact that * called with no args evaluates to 1:
(let[b(inc(*))j(+(* b b b)b)](+(* b j j j)j b b))
Using the same trick and doing string concatenation instead of arithmetic, the lowest I could get was 51 chars:
(let[n(*)t(+ n n)z(+)f(+ t t)](print(str t z n f)))

Ruby 1.9, 10 bytes
p 'ߞ'.ord

nice! 7 chars if you are in the irb command prompt 'ߞ'.ord

@EduardFlorinescu Thanks, I know.

These are 8 bytes in UTF8.

Python, 23
print ord("<DC3>")*ord("j")

<DC3> should be replaced with ASCII symbol 19 (device control 3).

SAS, 34 characters/bytes
data a;x=put(' ',hex.);put x;run;

That puts it to the log, it's 6 longer if you need it to the output window.  Note I'm not seeing the second character there; it is backwards-P, which is hex 14.
There should be a shorter solution with %sysfunc(putc(..., but I can't get that to work properly.

Edited: bash 81 chars!
Just for fun:
wc -c < <(echo {,}{,,}{,}{,,,}{,}{,,,}film dbugjkqstvxz{,}{,,} 'Happy New Year!')

there is no numbers, all letters are used and this print exactly:
2014

( This method could reasonably be used until 2016: by just adding one or two exclamation point after the wish:
   wc -c < <(echo {,}{,,}{,}{,,,}{,}{,,,}film dbugjkqstvxz{,}{,,} 'Happy New Year!!')
   2015

;-)

bash 27 chars
.;v=$?;echo $v$?${#v}$[v+v]

This will output:
bash: .: filename argument required
.: usage: . filename [arguments]
2014

Ok, this will generate some unwanted output, but 2014 is printed and is a valid token!
The two following sample are error free (a little longer but near golfed)
v=$(echo {V..v});echo $[${#v}#vu]
2014

or
printf -vv "%d" $?xfbc;echo $[v>>${#?}]
2014

or even:
echo $[$[$[${#?}$?-${#?}]$?>>${#?}]#Iy]
2014

Inspired by comment from GammaFunction:
echo $[$[a-a]xfbd>>${#?}]

You don't need to set a var in the 0xfbc solution: echo $[$?xfbc>>++j] works just fine (19 bytes).

Nice! Late to post but you could! Unfortunely you have to ensure $? to be 0 and this could not be reused... But impressive!

@GammaFunction Post edited ;-)

JavaScript 45
alert(parseInt('bbc','twentyonefour'.length))

Bash - 10 (or 8)
Well there have been a couple of answers that have been disqualified because they rely on the year. When golfing, one side goal is to see how close we can get to breaking the rules as currently written without breaking the letter of the rules (I include the clarifications by Joe Z in the 66 existing comments on the rules). The question very specifically states that I can not depend on 2014 being the current year. I instead rely on it being 8:14pm in my timezone.
date +%H%M

When I ran it, it output 2014 exactly, thus it satisfies it No, it has to be 2014 exactly. comment. (Due to context people seem to misread it as ... 2014 always, but that was not what was written, even if that were perhaps what was intended.) This lets me beat the current Bash record, at least until this loophole is closed. This interpretation may seem too cheaty since all the existing popular answers assume that the rules really meant always. Indeed some of them exploit this and export something that isn't exactly 2014, but instead contains 2014. I am fine with that interpretation too since Bash can do:
cat /*/*

This is a mere 8 characters, which will concatenates a bunch of files including /dev/urandom/, and it generally takes my machine under a minute to find 2014 in /dev/urandom. Although my rule twisting golfing code of honour won't let me pick this solution since it violates the letter of Joe Z's clarification, the only objection Joe Z  raised to the random approach in the 66 comments was that it was too long.  At 8 characters this answer is actually shorter than my rules-lawyer answer.

Normally entries that explore edge cases are popular. This is the only answer they obeys the letter of the rules that is down voted (all other down voted answers rely on year=2014). Anyone care to comment on what rubs them wrong about this answer?

It follows the letter of the rules in a way that's almost universally considered boring and unoriginal. Additionally, it was posted almost three months after the original problem, meaning that it's way back on the 4th page where barely anybody will see it, let alone vote it up.

For that matter, I've edited the question again to make my intentions for the problem even clearer. Is "independently of any external variables" good enough? Even then, this answer isn't the shortest in either category, so I still won't accept it.

Thanks. For the record, when I commented above this answer had been voted down to -2, so the context was more "would someone like to comment on why they are voting it down?" rather than "why isn't it upvoted?". It also wasn't intended as a criticism of Joe Z's question specification.

Python, 55 bytes (no math import and no char or unicode trickery!)
x=False;a=x**x;b=a+a;c=b+b;print c**c*(c+c)-b**(c+a)-b

Uses the fact that zero to the zeroth power is defined as one and False can be implicitly casted to 0. Hence a, b and c will contain 1, 2 and 4 respectively.

Arne, True implicitly casts to 1, so you can start with a=True and save 8 characters. Also, please specify Python 2.

Marbelous 14
CB
CE
CF
CD
~~

How it works
the first 4 lines are language literals, in hexadecimal. Their values are 203, 206, 207 and 205. They will fall down by one cell on each tick. If you perform an 8-bit binary not on those values (which is exactly what ~~ does) you get the following values: 52, 49, 48 and 50. These values happen to be the ascii values of 4, 1, 0 and 2 respectively. The literals then fall off the board which causes their corresponding ascii character to be printed to STDOUT.

C, 27 Bytes
main(){printf("%d",'\aÞ');}

Just a reminder that multi-character constants do exist :)
Alternatively, three bytes more:
main(){printf("%x",' i'-'U');}

int is not needed. 2. return doesn't print anything.

'^GÞ' (character codes 7 and 222) would require only one  multi-character constant.

My compiler does not agree :(

That's odd. 7 * 256 + 222 = 2014, so it should work. What do you get?

6178782, it seems like my compiler thinks of ^ as a separate character.

No, ^G is caret notation for BEL character; it's a single character with char code 7. You have to type the unprintable character somehow, depending on your OS and editor.

My compiler tells me it's '\aÞ'. Thanks anyway.

main is not needed, you can shorten it to f and save 3 bytes, it'll still be a valid submission.

Lua - 30 bytes
b=#"  "print(b..b-b..b/b..b+b)
# is the length operator, so b = 2.

Would b/b work?

@nooodl Good suggestion; indeed b/b can replace b-#" " and save 3 characters.

(Java, 553 bytes as .class, 112 bytes as it stands, 84 bytes after renaming the class to 'm' and removing whitespace.)
This probably isn't the kind of answer you're looking for, but there are a bunch of strings that share a hashcode of 2014.
public class make2014 {
    public static void main(String[] args){
        System.out.println("={".hashCode());
    }
}

Consider adding the language name (Java, I assume) and byte count like the other answers do.

Thanks, I forgot to do that. I didn't think my file size would matter much - the minimum file size of a .class is far greater than most of the attempts here.

I got the .java down to 84 bytes without changing functionality. I see other people shaved off 9 bytes by printing special characters to produce 2014 instead of using hashcodes... I kind of thought it would be cheating to do it like that, since you're really printing 2014 as a number in a different format. Oh well.

Xojo, 27 chars (all ASCII)
MsgBox Str(&hFBC/(&hC-&hA))

TinyMUSH, 16
We need more MUD language entries.
\encrypt($"#&,.)

C++ 30
main(){cout<<('&')*(']'-'(');}

Why do you need the brackets around '&'? Does main(){cout<<'&'*(']'-'(');} not work?

x86 machine code, 19 bytes
B8 3A 0E 2C 08 CD 10 2C 02 CD 10 04 01 CD 10 04 04 CD 10
Assembly code equivalent:
mov ax, 0E3Ah; ah = 0Eh (bios teletype), al = 3Ah (ascii semicolon)
sub al, 08h; ascii 2
int 10h

sub al, 02h; ascii 0
int 10h

add al, 01h; ascii 1
int 10h

add al, 04h; ascii 5
int 10h

Yeah, I know: it logs 2015 rather than 2014.
But seeing that this challenge is old and now the year is 2015, it seemed more appropriate to use the current year (it's my excuse for not "going home" :) )
Note: This was tested using DOSBOX

You have numbers in your source.

@ElliotA. Read the challenge: "without using any of the characters 0123456789"; numbers = characters representing numbers.

JavaScript (19)
Obvious cheating, but these expression ran in REPL print strings "2014" and "2015":
''+'ߞ'.charCodeAt() // 2014
''+'ߟ'.charCodeAt() //2015

TIL: .charCodeAt implicitly converts it's first argument to 0.

Vitsy, 5 4 Bytes, 3 Characters
When in Rome...
'Nߞ
Get the character with the value 2014 and then print it as a number. Simple.
More Interesting Version (12 10 9 Bytes):
"ca-^b-N-
My language supports hexadecimal, too. ;)
"         Capture the entire source as string by looping around the source.
 ca-      Push 2 to the stack
    ^     45^2
     b-   -11
       N  Output as a number.
        - Only here for character value 45.

Am I right that this language is newer than the question? Besides, I believe this question is already 'closed'(winner chosen, and even an edit at the start discouraging more replies)

@sanchises You're absolutely right - I'll pull my edit request. I'd still like to add it to the list, though, even if marked as a new language.. :D

Just put it under 'invalid' I guess, there's a section for that at the end.

Changed the edit suggestion. :P Forgot about the 'newer than question' thing, but I should've considered that. Thanks, @sanchises

Perl 5, 8 28 bytes

    say 38*53
Seems to do it.

Oh, without cheating ?
$z=ord("!")*ord("=");say++$z

D'oh.  I thought it was without any of those numbers.

Milky Way 1.0.0, 22 bytes
<^a:::+;:l+:>h<::++-<-

Explanation
<          <     <   # rotate the stack leftward
 ^                   # pop the TOS without outputting
  a                  # logical not on the TOS
   :::  :   ::       # duplicate the TOS
      +       ++     # push the sum the top two stack elements
       ;             # swap the top two stack elements
         >           # rotate the stack rightward
          h          # push the TOS to the power of the second stack element
                - -  # push the difference of the top two stack elements

The stack defaults to ["", 0].

Stack Visualization
["", 0]                # default stack

[0, ""]                # <
[0]                    # ^
[1]                    # a
[1, 1, 1, 1]           # :::
[1, 1, 2]              # +
[1, 2, 1]              # ;
[1, 2, 1, 1]           # :
[1, 2, 1, 10]          # l
[1, 2, 11]             # +
[1, 2, 11, 11]         # :
[11, 1, 2, 11]         # >
[11, 1, 2048]          # h
[1, 2048, 11]          # <
[1, 2048, 11, 11, 11]  # ::
[1, 2048, 33]          # ++
[1, 2015]              # -
[2015, 1]              # <
[2014]                 # -

By default, if nothing has been output manually, the bottom stack item is output on termination of the program.

Milky Way (current version), 8 bytes
XZ*W+U+!

Explanation
X         # push 20 to the stack
 Z        # push 100 to the stack
  *       # push the product of the TOS and STOS
   W      # push 10 to the stack
    + +   # push the sum of the TOS and STOS
     U    # push 4 to the stack
       !  # output the TOS

SMBF, 15 bytes
\x00 is a literal NUL byte. This program adds 5 to each of /,+- and prints.
<[+++++.<]\x00/,+-

Only my Python interpreter can accept non-printable ASCII. Change the data line to this, and substitute the code you want to run with the non-printable values escaped (on line 169):
data = bytearray(b'the above code goes here')

Retina, 24 bytes (newer than challenge)
Note the trailing space on lines 2 and 3. Language is newer than the challenge.

xx  x xxxx
+`(x)*
$#+

Try it online

Factor, 10 bytes
Not gonna beat my winning answer, but as a follow-on to the other "2014th Unicode char" answers:
CHAR: ߞ .

Prints 2014.

Base64, 9 Bytes
MjAxNA==
(You can decode it with: echo MjAxNA== |base64 -d)

Jelly, 10 bytes
⁹⁴×H_⁴Ḥ¤’’

Explanation:
⁹            Set the current value to 256.
 ⁴×          Multiply by 16. The current value is now 4096.
   H         Divide by 2. The current value is now 2048.
    _⁴Ḥ¤     Subtract by 16/2. The current value is now 2016.
        ’’   Decrement twice. The current value is now 2014.

Mathematica, 10 bytes
N[E,E^E^E]

Prints the decimal expansion of the number e to over 3.8 million decimal places. The first occurrence of 2014 in that decimal expansion starts at the 3180th decimal place.

VBA, 21 characters
?cells(,"BYL").column

Write and run the above code in the Immediate Window. Basically, the code converts column name BYL to its column index (2014).

For future reference, this may be rewritten as ?[BYL:BYL].Column

Actually, 15 bytes
This language was created long after this challenge was made, but I thought I'd still try my hand at it. This answer avoids all numerals, including Actually's ² for a*a. Golfing suggestions welcome. Try it online!
╜⌐u;*⌐úl¬¬τu;*-

Ungolfing
╜    Push register 0 (initialized to 0).
⌐u   Add 2 and increment. Returns 3.
;*   Duplicate and multiply. Equivalent to squaring. Returns 9.
⌐    Add 2 again. Returns 11.
úl   Pushes the lowercase alphabet and gets its length. Returns 26.
¬¬   Subtracts 2 twice. Returns 22.
τ    double(). Returns 44.
u    Increment. Returns 45.
;*   Square. Returns 2025.
-    Subtract. Returns 2025 - 11 == 2014.

Vim 8.0, 15 bytes
:h u
ggf:wywZZp

I didn't see a vim answer yet, so I figured I'd add one. This opens up a helpfile, so it is specifically vim 8.0, since it might not work with a future version that updates that file.

Clever! In my case :h followed by 9wywZZp works. So down to 10 symbols/bytes

@defhlt OK, good to know! Although technically that doesn't work because it uses the digit 9. You could do f:w in place of 9w for only one byte more. Which version of vim do you have?

Also, !!date +\%Y is 12 symbols (assuming you are reading this from 2014)

You are totally right! I use nvim 0.1.6-dev.

JavaScript, 81 76 bytes

l="length";alert(("hi"[l]<<"javascript"[l])-"wow"[l]*"hello death"[l]-true);

C, 24
(Using GCC-4.9.2)
f(){printf("%i",'Þ');}            (Properly formatted version on Pastebin)

IMPORTANT NOTE: There is a U+0007 in-front of the Þ character but stackexchange removes it for some reason. Make sure you edit it back in before compiling my code. Here is a pastebin of the code that does include the unicode characters needed.

SmileBASIC, 10 bytes
?&HFBC>>!.

&HFBC is hexadecimal for 2014*2, which is right shifted by not(0.0)
?ASC("ߞ") looks shorter, but it's actually the same length when saved in UTF-8, and about 100000x more boring.

√ å ı ¥ ® Ï Ø ¿, 8 bytes
TTX''_o

Try it online!
The rest of the code in the TIO link is the Python interpreter (because I can't be bothered to ask Dennis to add √ å ı ¥ ® Ï Ø ¿

Braingolf, 3 bytes
#ߞ

Try it online!
The ordinal of ߞ is 2014, # pushes the ordinal of the next character to the stack, and Braingolf implicitly outputs the last item on the stack.

Braingolf is your language, right?

@MDXF Indeed it is

Charcoal, 5 bytes
Ｉ²⁰¹⁴

Try it online!
Language was created after January 1, 2014, but as Charcoal uses the superindices ⁰¹²³⁴⁵⁶⁷⁸⁹ to represent the numbers, the answer is valid. :-)

Lua, 27 bytes
Should work in Lua 5.1, Lua 5.2, and Lua 5.3.
x="ɅɅ"print(x:byte()..#x)

Try it online!
This is mean to be saved with the UTF-8 encoding. The first byte of the string is 201, and its length is four. Lua is mostly encoding agnostic, so as long as these things are true in whatever encoding, it works.
With only ASCII, 28 bytes:
x=""io.write(x:byte(y,#x))

Note: the string must contain ASCII 20 and ASCII 14 (which are not printable characters). y here is an undefined variable, so it is nil, which byte defaults to 1 in the first parameter.

Lua, 29 bytes
b=("").byte print(b"?"..b"?")

NB: the two question marks are substitutes for characters that are not appearing properly in the post. See the tio link below for proof.
Try it online!
As an interesting point, although this is not the case with Lua, a language with an implementation of pi to at least 3137 digits would be able to print pi and find '2014' at digits 3133-3136!

Python, 41 bytes
print(int(ord("j")/len("aa") * ord("&")))

Try it online!

Welcome to the site! The aim of this question (and code golf in general) is to get the shortest functional code. You can remove the spaces around the * to start with.

PHP, 25 bytes
<?=IntlChar::ord("ߞ");?>

This requires the IntlChar library; and that´s a lot younger than the question. Nice idea, though.

Thanks. Can be solved without using IntlChar library with preg_replace('#\D#','',json_encode('—')). Pick your number from a unicode table :-)

Pyt, 9 bytes
π!⬡⁻⁻⁻△⁻⁻

Try it online!

LibreLogo (Paper Format: Tabloid), 55 bytes

The document must be in Tabloid Format for this method to work.

Code:
print ''.join(set(str(pagesize.pop()))).replace('.','')

Result:

Explanation:
pagesize                                           ; Returns [792.0, 1224.0] (Tabloid Format)
pagesize.pop()                                     ; Returns 1224.0
str(pagesize.pop())                                ; Returns "1224.0"
set(str(pagesize.pop()))                           ; Returns {u'2', u'0', u'1', u'.', u'4'}
''.join(set(str(pagesize.pop())))                  ; Returns "201.4"
''.join(set(str(pagesize.pop()))).replace('.','')  ; Returns "2014"

Canvas, 7 bytes
ＡＡ＋ｑ⁷┤ｔ

Try it online!
ＡＡ       | Push 10 to stack twice
   ＋      | Add top two items in stack
    ｑ     | Print top item in stack on current line
      ⁷    | Push 16 to stack
       ┤   | Decrement top item in stack twice
        ｔ | Print top item in stack on current line, disable implicit output

Perl 5 + Perligata, 25 bytes
Byte count uses the new rule that command-line switches for choosing appropriate modules don't count.
perl -MLingua::Romana::Perligata -e'MMXIVimum egresso scribe.'

Tested in Strawberry 5.26.0 with Perligata 0.601 (with " instead of ' in the above code).

T-SQL, 32 bytes
Based on the idea of Steve Matthews, but avoiding unprintable characters:
SELECT -~(ASCII('=')*ASCII('!'))

Alternative (but rather obvious) solution, 23 bytes:
PRINT UNICODE(N'ߞ')

jshell, 5 bytes
How to make Java an useful code golf language? Simple, just remove the need to declare classes, methods, make things automatically print and so on.
+'ߞ'

Stax, 4 bytes
ü◘♥┐

Run and debug (ha) it online!
Obligatory Stax answer. This unpacks to 2014. Implicit print.
I haven't yet found a shorter version.

C# (.NET Core), 42 bytes
System.Console.Write((int)'ϯ'+(int)'ϯ');

// Btw. it's now 2018 ^^ so
System.Console.Write((int)'ϱ' + (int)'ϱ');

Try it online!

CJam, 2 bytes
KE

K is predefined as 20, E is predefined as 14. The stack gets automatically printed after the program.
Try it online!

C#, 28 bytes
You don't need to cast to an int as Hille does, so it becomes
Console.WriteLine('ϱ'+'ϱ')

Symbolic Python, 31 bytes
_=-~-~-~(_==_)
_=`""`[_::_]

Try it online!
Symbolic Python bans numbers anyway.
Explanation:
_=-~-~-~(_==_)   # Set _ to 4
_=               # Set _ to
  `""`         # From the representation of some unprintables
                    # Which is '\x12\x10\x11\x14'
           [_::_]   # From the 4th character, take every 4th character
                    # Output the contents of _ implicitly

A more interesting solution at 34 bytes:
_=-~(_==_)
_=`_`+`_-_`+`_/_`+`_*_`

Try it online!
_=-~(_==_)               # Set _ to 2
_=`_`+`_-_`+`_/_`+`_*_`
# '2'+'2-2'+'2/2'+'2*2' = '2'+'0'+'1'+'4' = '2014'

Crazy that someone else is actually using this stupid language I made, haha

@, 9 chars
-Σ*{~~}82

Explanation (syntactically invalid)
-       2  Subtract the result by 2(2014)
 Σ         Summation of all ASCII codes in the string(2016)
  *    8   Duplicate the string 8 times
   {~~}    Define the string "~~"

There is no Try It Online for @.
If the current year is 2016, it would be perfect(7 chars):
Σ*{~~}8

><>, 7 bytes
aa+nen;

You can try it on The Online ><> interpreter.
Explanation
a        # push 10
 a       # push 10
  +      # push sum of top two values (10 + 10 = 20)
   n     # output top value as number
    e    # push 14
     n   # output top value as number
      ;  # halt program

This doesn't print the following:
20
14

because n outputs the top value as a number without a newline.
Alternate Solution (9 bytes)
This one actually puts 2014 on the stack as a single number and outputs it.
cbde**+n;

I basically just tried random values for this one :P

Javascript, 45 characters
_=> new Date().getFullYear()-new Date().getDate()

Very Temporary!!
Another Day, another very Temporary answer :-)
f=new Date().getFullYear()-new Date().getDay()

Welcome to Code Golf! Submissions should be either a full program or a function, so I would recommend prepending _=> to your solution to turn it into an anonymous function.

You could possibly remove the space between => and new to save a byte

Triangular, 12 bytes
tE**%Cdd.`>/

Try it online!
Y'know, I was just about to post a 14-byte solution; luckily I went back to double-check and had a hunch.
Ungolfed:
    t
   E *
  * % C
 d d . `
> /

-- Actual Execution Order --

t*C`          The first multiplication does nothing.
              12 is pushed twice, then direction changes and both are multiplied to get 144.

E*d>/d%       Push 14, then multiply 14*144=2016. Decrement twice and print.

The 14-byte I was going to post:
t*CE.`*...>dd%

Wren, 27 bytes
Just like most other answers, convert to code points and print. However this uses control characters in the source code.
System.printAll("".bytes)

Try it online!
Explanation
                ""        // An unprintable chain containing 0x14 (20 in decimal) and 0x0e (14 in decimal)
                  .bytes  // Convert them to a list of their decimal codes [20, 14]
System.printAll(        ) // Print them all without a separator (2014)

ABC, 11 bytes
aacncacaaac

Explanation
a increments the accumulator, c outputs it and n sets it to 0.

W j, 4 bytes
ë‘"C

Explanation
ë‘"  % Push a list
   C % Convert to list of ord codes [20,14]

flag:j % Join without a separator
```

MathGolf, 2 bytes
ID

Try it online!
Explanation
I   # push 20
 D  # push 14
    # implicit output

Spice, 145 97 bytes
Improvement using multiplication to get the 9 from 2*2*2+1, as we already had those values.
;t;z;o;n;i@LEN i z;ADD i i i;LEN i o;PUT i i i;LEN i t;MUL t t i;MUL t i i;ADD o i n;OUT t z o n;

Un-golfed explanation
;t;z;o;n;i@ - Declare vars
LEN i z;    - Get the length of [], 0, and store in z
ADD i i i;  - Adding implicitly uses the first element, or 0 if there is none, so we insert 0 into i
LEN i o;    - Store length of [0], 1, in o
PUT i i i;  - Insert 0th element of i into i at position i[0] (we're increasing the array size)
LEN i t;    - Store length of [0, 0], 2, into t
MUL t t i;  - Multiply t, 2, by itself and store in i
MUL t i i;  - Multiply t, 2, by i, 4, and store in i
ADD o i n;  - Add o, 1, to i, 8 and store in n
OUT t z o n;- Write to console - "[2] [0] [1] [9]"

Original
;t;z;o;n;i@LEN i z;ADD i i i;LEN i o;PUT i i i;LEN i t;PUT i i i;PUT i i i;PUT i i i;PUT i i i;PUT i i i;PUT i i i;PUT i i i;LEN i n;OUT t z o n;

Un-golfed explanation
In Spice, all variables are double arrays. Importantly, variables that have no assigned value are either treated as an empty list [] or 0 depending on the operation. The built-in LEN will give the length of an array and we can therefore produce numbers:
;t;z;o;n;i@ - Declare vars
LEN i z;    - Get the length of [], 0, and store in z
ADD i i i;  - Adding implicitly uses the first element, or 0 if there is none, so we insert 0 into i
LEN i o;    - Store length of [0], 1, in o
PUT i i i;  - Insert 0th element of i into i at position i[0] (we're increasing the array size)
LEN i t;    - Store length of [0, 0], 2, into t
PUT i i i;  - Now repeat until there are 9 elements...
PUT i i i;
PUT i i i;
PUT i i i;
PUT i i i;
PUT i i i;
PUT i i i;
LEN i n;     - ... and store in n
OUT t z o n; - Write to console - "[2] [0] [1] [9]"

For the original 2014 version, you save bytes for less PUTs - 95 bytes. So this solution will improve next year!

FEU, 69 bytes
a/abcdeghij
m/a/bb/b/cc/c/dd/d/ee/e/ff/f/gg/g/hh/h/ii/i/jj/j/kk/g
U/k

Try it online!

Excel, 12
=UNICODE("ߞ

Second best I could do was =ARABIC("MMXIV at 14 bytes.

MAWP, 30 28 bytes
!!+!!!!++!*+/!+!!+!+!!++++*:

Try it!
This will be fun to golf.
This is longer than @Lyxal 's answer, but outputs only one time as one number.

15 bytes

Vyxal, 10 3 bytes
»÷∩

Try it Online!
Simply the compressed number 2014

Can't be bothered to download the interpreter, but does !!!!!''.$... work for 11 bytes?

Poetic, 112 bytes
two`s a bad thing
using a two?o,hardly!i am using a poem
a numeric poem?oh,clearly,but a limited one for certain

Try it online!

DROL, 13 bytes
ziill<ukl<dfo
I wanted to add a language that wasn't already included...
DROL is a limited instruction assembler-like language with only two registers as storage.  The language does include numbers, so it think it qualifies for this question.  But numbers are only used for loop length indicators, as well as being used as the name of some instructions.  It's described on the Esolang Wiki DROL page.
Here's a rundown of what the code does:
z              set R1=0
 ii            increment R1 by 1 twice   (2)
   ll          square R1 three two      (16)
     <         shift left R1            (32)
      u        set R2=R1                (32)
       k       increment R2             (33)
        l      square R1              (1024)
         <     shift left R1          (2048)
          d    decrement R1           (2047)
           f   subtract R1 = R1 - R2  (2014)
            o  print R1 as an integer

ARM assembly, 94 85 bytes (28 bytes compiled)
Textual assembly for ARM mode.
s:subs sb,sb
adc sb,sb
add sb,sb
lsl sl,sb,sb
add sb,sl,lsl sb
rsb sb,sl,lsl sl
bx lr

A function which returns 2014 in sb (r9)
Clobbers sl (r10) and sb (r9).
Expanded version:
        // It feels so empty here...
        .globl s
s:
        // use the quirky way subs affects the flags
        // to set r9 to 1
        subs    r9, r9, r9
        adc     r9, r9, r9
        // double r9 by adding it to itself
        // lsl works just as well
        // r9 = 2
        add     r9, r9, r9
        // r10 = r9 << r9
        // r10 = 2 << 2
        // r10 = 8
        lsl     r10, r9, r9
        // r9 = r9 + (r10 << r9)
        // r9 = 2 + (8 << 2)
        // r9 = 2 + 32
        // r9 = 34
        add     r9, r9, r10, lsl r9
        // r9 = (r10 << r10) - r9
        // r9 = (8 << 8) - 34
        // r9 = 2048 - 34
        // r9 = 2014
        rsb     r9, r9, r10, lsl r10
        // Return
        bx      lr

This uses the same idea as my Phooey answer, of generating 2048, then subtracting 34. While I do have access to push and pop, ARM isn't a stack machine. It is a register machine. Additionally, we have lsl for shifting left which makes a few cases easier.
It is yet another rare case where the inverted carry flag on ARM is useful for something other than 64-bit subtraction, as it allows us to set a register to 1 when combined with adc.
Additionally, this only works in ARM mode: Thumb-2 did not show the return of shifting by register (which kinda was a dumb waste of encoding bits)
It uses the classic register names instead of the format which is r[0-15].

Python 3, 16 bytes
print(ord('ߞ'))

Try it online!

Kotlin, 42 bytes
fun main()=print("ް.".map{it.code}.sum())

I used U+1968 which is ް  and a . which is U+46.

Phooey, 16 bytes
=@+$i>$i=$i<@+$i

Try it online!
                 # Stk  Tape
=@+$i>$i=$i<@+$i # (0) >0  0   (initial state)
=                # (0) >1  0   tape = stack == tape to get 1
 @               #  1  >1  0   push to stack
  +              # (0) >2  0   pop; add stack to tape
   $i            # (0) >2  0   print tape as integer (2)
     >$i         # (0)  2 >0   move tape ptr right, print (0)
        =$i      # (0)  2 >1   same as above to get 1 again, print (1)
           <@    #  2  >2  1   move back, push
             +$i # (0) >4  1   add stack to tape, print (4)

Thank goodness for Phooey's = operator. This would be impossible in Foo.

Phooey, 23 19 bytes
This version actually generates the number 2014 instead of printing 2,0,1, and 4.
=@+@@@+@**@@*@+--$i

Try it online!
                    # stack       | tape
=@+@@@+@**@@*@+--$i #         (0) |    0  initial state
=                   #         (0) |    1  tape = tape == pop() (to get 1)
 @+                 #         (0) |    2  double by adding to self
   @@               #    2     2  |    2  push two copies to the stack
     @+             #    2     2  |    4  double
       @*           #    2     2  |   16  square by multiplying by self
         *          #          2  |   32  multiply by 2
          @         #    2    32  |   32  push 32 for later
           @*       #    2    32  | 1024  square
             @+     #    2    32  | 2048  double
               -    #          2  | 2016  pop and subtract
                -   #         (0) | 2014  pop and subtract
                 $i #         (0) | 2014  print 2014

Grok, 13 bytes
iNH`I:P-zP-zq

Alternate 13 Byte solution:
i:H:N`-Yx-zZq

Branch, 11 bytes
}}#/#}#^}}#

Try it on the online Branch interpreter!
Branch does have numbers in it. } is increment and { is decrement. # outputs as number. / goes to the left child, which is automatically initialized to 0, which is shorter than {{. Finally, ^ goes to the parent, which is 2 when that command is called. Actually, since the current node is 1, and the parent is 2, we could do ^}} or }}} to get 4.
An alternative solution that produces 2014 on the tree itself instead of outputting it character by character:
Branch, 50 bytes
}}^\}}}}}^*^\}}^*{^\}}^*^\/}}}}}}^\}}}}}}}}}^*{^*#

Try it on the online Branch interpreter!

Deadfish~, 12 bytes
iioddoioiiio

Try it online!
This, this is surprisingly short!

save bytes by squaring instead of incrementing twice at the end

This can just be vanilla deadfish, since you don't use ch{}(). Also, iiio can be replaced with iso for 11 bytes.

And you can get the number 2014: iiisddsddddsdddddddddddo but that's a bit long.

PICO-8, 24 18 bytes
?ord("⁙")*ord("j")

ord() gets the numerical index of a character; in P8SCII, ⁙ is at index 19 and j is at index 106, creating the equation \$19*106=2014\$.

Python 3 (15 Bytes)
print(ord('ߞ'))

Python 3 (24 Bytes)
print(ord('޴')+ord('*'))

JavaScript (28 Bytes)
-~!+[]+[+[]]+-~[]+-~!''-~!''

I'm pretty sure this is a dupe of a large number of existing answers: https://codegolf.stackexchange.com/search?q=inquestion%3A17005+Python

C++ (gcc), 33 bytes
main(){std::cout<<('j'-'W')*'j';}

Try it online!

Should #include <iostream> be included as byte count?

Might as well go for putting \x13 in your source! -6 bytes

@tsh I agree the include should be part of it. In the same vein you could just switch to C

Pascal, 76 B
See also Delphi.
This program requires a processor supporting ISO standard 10206 “Extended Pascal”, specifically constant definitions may be expressions (ord(true) is an expression).
program p(output);const I=ord(true);begin write(I+I,I-I:i,I:i,I+I+I+I:i)end.

If ord(maxChar) ≥ 2014 we can use just one character in the 43 character program
program q(output);begin write(ord('X'))end.

where X needs to be substituted by the implementation-defined char value having the ordinal value 2014.

For reference, an Extended Pascal program printing the current year looks like this:
program t(output); var t: timeStamp; begin getTimeStamp(t); writeLn(t.year) end.

Go, 56 bytes
func f(){for _,c:=range`-+,/`{print(string(c+'F'-'A'))}}

Attempt This Online!
Alternative to that other Go answer. Prints to STDERR.

tinylisp 2, 5 bytes
(h"ߞ

Try it at Replit!
Explanation
The same character-code solution as everybody else. h is short for head, which, when given a string, returns the character code of its first character. The string quote and the function-call parenthesis both auto-close at the end of the line/program.
Non-string solution, 43 bytes
(d T(+(*)(*
(d X(* T T T T T
(-(* X X T)X T

Here, we take advantage of the fact that tinylisp 2's multiplication builtin * is variadic: it can take any number of arguments, including zero. Without arguments, its return value is the product of an empty list, or \$1\$.
(d T(+(*)(*)))

Define T as the sum of \$1\$ and \$1\$, or \$2\$.
(d X(* T T T T T

Define X as the product of five \$2\$s, or \$32\$.
(-(* X X T)X T

Output the product of \$32\$ and \$32\$ and \$2\$, minus \$32\$, minus \$2\$ (the - builtin is also variadic): \$(32 \cdot 32 \cdot 2) - 32 - 2 = 2014\$.

Logically, 68 bytes.
@w:a,b,c;WRITE(h,a,b,c,l,h,h,l,l)()
@M w(l,h)()w()()w(h)()w(l,l,h)()

Consists of two gate definitions. w writes the 3 passed bits in the form of 00110cba, and then M has 4 w gates for each of the digits.
Whilst logically can use numbers for both argument unrolling and high-low states, it can be avoided pretty easily.

Thunno 2, 3 bytes
«¬ẓ

Attempt This Online! Simple compressed integer.
Thunno 2, 4 bytes (UTF-8)
'ߞC

Attempt This Online! Charcode.
Thunno, 4 bytes (UTF-8)
'ߞO

Attempt This Online! Basically the same so I won't create another answer for it.

Thunno 2 HHHHHHHHHHHHHHHHHHHHTs, 1 byte: 4

@Jacob you goofy - 4 is a digit and no digits allowed in the answer. You too noodles for golf.

@lyxal oops fixed it: Thunno 2, {"Z"*2014}l, 0 bytes.

(and, in case it’s not clear, those are joke submissions and obviously unfair)

@Jacob lol, nice ones

!+~%, 8 bytes
++!+~%+!

Stumbled over this challenge and sorry, could not resist. I promise this will not happen again. (-;
An explanation, to make this look serious:

++ twice adds 7 to the initial 6, so we end up at 20, that gets printe by !
to the accumulator resetted to 6, + adds 7, and the ~ reverses the result from 13 to 31
% does a modulo from the number with a leading 1, so 131 mod 31 happens to be 7.
the + adds another 7, and ! prints the 14 behind the 20.

morsecco: 24 bytes
Now a more serious answer:
. -----.----. -.- -. ---

Morsecco codes numbers as binaries with dots and dashes (you could also write it as morse code, but each digit takes 6 bytes, resulting in a much longer code), so . -----.----. Enters 2014 on the stack, -.- -. Konverts this to a decimal Number which gets Output by ---.

Vyxal 3, 3 bytes
~¥ᴴ

Try it Online!
because of how the encoding works, vyxal 3 takes 4 bytes to use a base 252 encoding, but you can shave a byte because 2014 is only 2 bytes in base 255 and thus you can use the ~ element which converts 2 elements on the codepage to base 255

Objective C
NSDateFormatter *formatter = [[NSDateFormatter alloc] init];
[formatter setDateFormat:@"yyyy"];
NSLog(@"%@",[formatter stringFromDate:[NSDate date]]);

From the comments: Gelatin: “Is it acceptable to use the current year?” Joe Z.: “No, it has to be 2014 exactly.”

Because the question is a code-golf question, please add the character count.

What happened to you, Smalltalk ?  You look...different.

NSLog(@"%i",'&'*('F'-'A'));

@AlbertRenshaw that just prints 190.

@Cœur sorry meant this NSLog(@"%i",'&'*('V'-'!'));

Should be noted that OP's source code just prints 2018 ;)

You never said we couldn't put it on an external page!
Javascript - 19 Chars
location='//x.vu/u'

PS.  It took 2 tries to get a shortened URL without numbers :P

Yes, but sadly your code is far from being the shortest. :P

@Joe Z. Right, you could include an HTML script src instead :P

location.href.match(/-(\d+)-/).pop()

location='http://goo.gl/miVwHe' would be shorter in my opinion.

@xfix Thanks for the tip, I even removed 'http:' :)

You can take out those two slashes at the start too. ;)

@Trimsty I wish, but unfortunately not ;(

@TheWobbuffet Hm. That's really weird, I really thought it'd work! Sorry.

C, 29 bytes (Emacs) / 40 bytes (VI)
Before creating program, execute the shell command
stty -ixon      ### 11 bytes, with the newline

N.B this is required for most terminal editors, but is not required for emacs
then write this code:
main(){printf("%d",'j'*'^S');}

N.B. '^S' is a single XOFF character e.g. in VI use control-V, control-S; in emacs use control-Q, control-S; emacs turns off flow control by default.
I assume the comment [You can't type ^S in a Linux terminal] below left off an implied [without the stty command or equivalent e.g. emacs default behavior upon startup], otherwise that comment is inaccurate.
And here is the requested hexdump:
$ od -a -t x1 x.c
0000000   m   a   i   n   (   )   {   p   r   i   n   t   f   (   "   %
         6d  61  69  6e  28  29  7b  70  72  69  6e  74  66  28  22  25
0000020   d   "   ,   '   j   '   *   ' dc3   '   )   ;   }  nl
         64  22  2c  27  6a  27  2a  27  13  27  29  3b  7d  0a

This isn't a valid answer. You have to count the shell command also, since it's a non-standard configuration. Also, if you have unprintables in your code, you should include a hexdump, for extra clarity.

You can't type ^S in a Linux terminal. -1

There is more than one way to ^Skin a cat, so these comments are incorrect as written; I clarified my post to address the confusion these folks (and others) may be having.  It's still 29 if you use emacs as an editor.  MD XF:  how do you enter your [U+07DE] character into C-source; also, my compiler throws an error when replacing [main()] with [f()] (excluding the obvious gcc -Df=main hack), but if there is a compiler that accepts [f()], then all C source scores drop by 3.  So you need to bump your score by 3, or reduce all other C solutions.

GTB, 38
π/π→A:A++B:A+A→C:C*C→D~B*C*C+B*C-C-A-A

Compile assuming : at front

π is 2 bytes and → is 3, bringing the byte count to 49.

@JoeZ. All characters at http://tibasicdev.wikidot.com/83lgfont are 1 byte.

C# (56 characters)
Class P{static void Main(){Console.Write(','*','+'N');}}

This doesn't print anything.

26 characters = 4 bytes?? I don't think so...

Sorry. I just included the logic only. Now I included the whole program.

You need System.Console to use Console.

using System; will be on the header. So....

Perl, 24 bytes
print-ord(A)+ord for U,O

Pure ascii, no nonprinting characters or utf8-only characters used.
Uses the 21st and 15th letters of the alphabet to print 20, 14.

BAT (windows batch), 7 characters
echo %*

save as a.bat and invoke as a 2014

is there any restriction on using command line parameters? – Einacio
I don't think so, but echo $1 still has a number in it. – Joe Z.

not a so valid answer, but noone was posting an answer in this wonderful language

Does BAT have an equivalent to sh’s $@ for “all parameters”?

@ZevEisenberg that's what i used, altought it seems someone didn't like it

PHP, 22 bytes
echo ord('').ord('')

Edit
SE is stripping the characters. Working version here: http://codepad.org/unzjXNY2.

this produces "00" when I run it. Please explain.

SE is stripping the characters. Added a codepad example.

You can swap out echo  with <?=, to make it 2 characters shorter.

PHP (27) - not shortest ASCII PHP but more readable
<?=hexdec($e=hexdec(E)).$e;

Bash, 15 bytes
echo "^T "|od -x

(^T is one character.)

How about: printf %x \'— 13 characters.

Bat (Windows batch), 13 bytes (characters)
echo.^T |od -x

(^T is one character.)

PYTHON
print(str(len('Happy new year to me')) + str(len('Happy new year')))

This has a 1 in it.

oh whoops let me fix it

Must str('Happy new year') be wroted as str(len('Happy new year'))?

sorry. I fixed it.

k 11 & 17 chars
this 11-char one is from a colleague
@:[:']*.(.)

this 17-char one is the best i could come up with on my own; it's a variant on ASCII abuse:
.,/$-/"i"$$`zz`fl

and just for interest, here are a few others of mine:
."c"$"RPQT"-"e"$" " / ascii (only 2.x)
."c"$-/"e"$("RPQT";" ") / ascii (all versions)

-_-(s*(exp acos@-`=`)xexp x)-(s xexp s:x*x)%x:+/``=`` / port of David Carraher's solution above

.,/$#:'(``;();`;````) / another approach

@David Carraher my latest improvements on my port of your algo--two implementations, both 39 chars:
-_-(x*p*p:exp acos@-#`)-f*x*x*x:f*f:#``
f+(x*_p*p:exp acos@-#`)-f*x*x*x:f*f:#``

(i don't have the rep to comment on your post yet)

Delphi (26bytes & 26 chars)
ord('-')*ord('/')-ord('e')

Ascii values
  - : 45
  / : 47
  e : 101
  45*47 = 2115 - 101 = 2014

The specifications say “make a program that prints”. This code excerpt is merely an expression that’ll evaluate to 2014. It’s not a program nor will it print anything. Complete Pascal solution.

e-TeX, 21 bytes
\the\numexpr`*`j\bye
It contains an invisible control character with code 19 (0x13) before the asterisk. A version with printable ASCII characters needs two more bytes:
\the\numexpr`^^S*`j\bye

In TeX ` takes the character code of the next token:

[0x13] (^^S): 19
j: 106

\numexpr calculates: 19 * 106 = 2014
The result is a DVI file with "2014" on the first page.
Variant with 2014 as page number:
\pageno\numexpr`^^S*`j~\bye

(25 bytes, if ^^S is replaced by the byte with character code 19).

Java without the weird unicode charas @ 115
enum A{A;public static void main(String[]z){int a=A.ordinal(),b=a++;System.out.print(""+(a<<a)+b+a+(a<<(a<<a)));}}

enumerators are pretty handy :)

Python, 8 characters
ord('ߞ')

I think this ought to be valid :P
Ok, if you want me to use print(),
15 characters
print(ord('ߞ'))

Microscript, 10 bytes
Noncompeting, language postdates the question..
'js'(s';-*

Explanation: 106*(59-40), using the language's equivalent of character literals. I'd do 53*38, but the character corresponding to 53... is the digit 5. And the character corresponding to 19 is, of course, nonprintable, so that wouldn't work.

Cardinal, 14 chars
%n=*+.-d++.
 d

The pointer starts at %, reads in the character ‘d’(ASCII 100), creates a duplicate as inactive value (the bottom of the stack), adds active and inactive values, resulting in 200, adds 1, prints out the result 201, subtracts one, divides by inactive value (100), resulting in 2, adds two, prints out 4.
Cardinal pointer stacks can only carry values up to 255 (OEM 437 range), everything above leads to a wrap-around.
0             100             100             200             201
0               0             100             100             100
>n=*+.-d++.    %>=*+.-d++.    %n>*+.-d++.    %n=>+.-d++.    %n=*>.-d++.
 d              d              d              d              d
———————————————————————————————————————————————————————————————————————
   201             200               2               3               4
   100             100             100             100             100
%n=*+>-d++.    %n=*+.>d++.    %n=*+.->++.    %n=*+.-d>+.    %n=*+.-d+>.

print:"201"
———————————————————————————————————————————————————————————————————————
          4
        100
%n=*+.-d++>

  print:"4"

Result:
Executing program..

2014

Execution complete.

C, 24 / 68
main(){printf("MMXIV");}

Haha, Roman Numerals For The Win!

Or, for real:
main(){printf("%i%i%i%i",strlen("aa"),nil,strlen("a"),sizeof(int));}

q, 20 17 bytes
{x+y*z}."j"$"$+."

TeaScript, 7 bytes
'ߞ'c()

Takes the char code of the character with a char code of 2014

Groovy, 48 bytes
For 2015 change cabe to cabf...

n={it.each{print(((int)it)-(int)'a')}}
n('cabe')

JavaScript, 17 Bytes

atob("MjAxNA==")

Swift 2.2, 28 bytes
print("\(ENOTDIR)\(EFAULT)")

Falling back on Darwin/glibc error codes from errno.h.

Mathematica, 16 bytes
FromDigits@"JAE"

C (gcc 5.3.1), 49 bytes
Pure arithmetic! \o/
main(a){printf("%d",(++a<<(a<<a|a))-(a<<a+a)-a);}

Undefined behaviours.
Specifically, use this compiler (languages are defined by implementation).
Without undefined behaviour (which would work on all compilers), 51 bytes
main(a){a++;printf("%d",(a<<(a<<a|a))-(a<<a+a)-a);}

Explanation
Basically 2048 - 32 - 2, constructed using powers of 2.

QBIC, 10 bytes - Not Competing
?z^s*r+z+t

Calculates 2014 from the pre-initialised variables q-z (1-10) and prints it. Development of QBIC started some two years after 2014...
Alternative 12-byte version:
?(u*y)^r-z-q

Pushy, 6 bytes
`��`j#

Contains control characters, so here's a hexdump:
0000-0006:  60 14 0e 60 6a 23

The first unprintable is the literal DC4 byte (\x14), and the second is SHIFT-OUT (\x0e).
First these bytes are pushed as charcodes, so the stack is [20, 14]. The j operator concatenates these and # outputs the result: "2014".

10-byte solution:
`<:;>`KT-"

Uses char-code manipulation and the builtin T (10).
`<:;>`       Push string as char-codes: [60, 58, 59, 62]
     KT-     Take 10 from each: [50, 48, 49, 52]
        "    Interpret as char-codes and print: results in "2014"

Pyke, 3 bytes, noncompetitive
w߾

Try it here!
Where ߾ is 0xDFBE
Loads ord(0xDFBD-0x20) as an integer and implicit prints it

How does it work?

@Pavel explanation added

tcl, 16
puts [scan ߞ %c]

Can be seen on: http://rextester.com/live/SVXB29034

Groovy, casting to int, 8 bytes
(int)'ߞ'

C, 25 bytes
f(){printf("%d",L'ߞ');}

24 characters, but one character is UTF-8 encoded. Still the shortest C answer! How it works:
U+07DE ߞ NKO LETTER KA

7DE in decimal is 2014.

Python 3, 16 Bytes :
print(ord('ߞ'))

2017 version, 16 Bytes :
print(ord('ߡ'))

Whitespace, 19 bytes
Visible representation:
SSSTTTTTSTTTTSNTNST

Just pushes 2014 onto the stack and prints it. Whitespace's lack of any visible characters makes this pretty easy.
Valid as numbers are completely valid tokens in whitespace, they just don't do anything.

Common Lisp, 15 characters
(char-code #\ߞ)

Try it online!

Ruby (8 bytes, 7 chars)
'ߞ'.ord

The question mark is ߞ represented in two bytes UTF8.

Duplicate of O-I's Ruby solution posted 3.5 years earlier.

@manatwork Duplicate answers are allowed.

I didn't said they aren't. Neither downvoted or flagged. (Just for the record, I never agreed with that rule. And never will.) But I think would be more enjoyable to avoid duplicated solutions. Especially on a question that already has 263 answers.

Hm... hard to find the old solutions. But actually his solution is 9 bytes because the codepoints representations need 2 bytes.

Btw: Having seen the solution of @o-i : ?ߞ.ord would then be only 7 bytes, right?

Initially O-I's solution was also 'ߞ'.ord, but then be updated because the requirement says to “prints the number 2014”. The general rule (which can be overridden by each challenge) is that solutions must handle input and output either themselves explicitly or benefit the interpreter's service if it has such thing like ruby's -n or -p. Code that expects input to be readily available in the memory or just leaves the value they produce in the memory are called snippets and generally are not accepted as solutions.

BTW, looks like neither Timtech found O-I's solution and posted his own after just 1 week. I assume there were considerably fewer solutions that time.

Clojure - 35 characters ASCII only
(print(-(*(int\#)(int\<))(int\V)))

Based on True Soft's answer

Implicit, 3 bytes
`ߞ

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     implicit push command
`    character
 ߞ   U+07DE (2014 in decimal)
     implicit integer output

Aceto, 10 bytes
IIppIpPiIp

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Uses int(Pi) to it's advantage
11 bytes (previous answer)
IIppIpIIIIp

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Simply increments from zero 2, 0, 1, and 4 times, then prints each time
Fun Version
Piddd++DsJiPidIIJi*p

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Uses int(Pi) to base equations off of
Lame Version
'ߞop

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Converts the codepoint to number, which is 2014

Brainf*ck
>++++++++++[<+++++>-]<.--.+.+++.

More readable version:
> Move pointer pos over
++++++++++ Add 10
[ Start loop
<+++++>- Add 5, move over & subtract 1
] End loop
<.--.+.+++. Do some other stuff

Yes, I know this is an old post, but I'm really bored and want to get enough rep to comment because I don't have access to my other account.

Welcome to PPC...Jk

Brainfuck, 23 Bytes
-[>+<-----]>-.--.+.+++.

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Triangularity, 18 bytes
.. ..
."ߞ".
o

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Pyt, 16 bytes
áŁ!⁺²Đ⁺⁺!₀⁻⁻⇹ᴇ+₅

Explanation:
á                               Push stack into list (pushes empty list)
 Ł                              Get length of top of stack (0)
  !                             0!=1
   ⁺²                           (1+1)^2=4
     Đ                          Duplicate top of stack
      ⁺⁺!                       (4+2)!=720
         ₀⁻⁻                    720/10-2=70
            ⇹                   Swap top two items on stack (4 is on top)
             ᴇ                  10^4=10000
              +₅                (10000+70)/5=2014

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SNOBOL4 (CSNOBOL4), 42 bytes
	&UCASE 'N' @X ARB 'T' @Y
	OUTPUT =Y X
END

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@ assigns the index of the match and  (space) concatenates them.
I know, it's 2018, so I should go home...

Objective-C, 27
NSLog(@"%i",'&'*('V'-'!'));

Python 3, 16 bytes
lambda:ord('ߞ')

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x86, 15 bytes
No numbers in the source code. Returns in eax. Uses (252*4 - 1) * 2.
   0:   31 c0                   xor    %eax,%eax
   2:   fe c8                   dec    %al
   4:   48                      dec    %eax
   5:   48                      dec    %eax
   6:   48                      dec    %eax
   7:   d1 e0                   shl    %eax
   9:   d1 e0                   shl    %eax
   b:   48                      dec    %eax
   c:   d1 e0                   shl    %eax
   e:   c3                      ret

If we permit numbers in the source but not binary, we have mov $2014,%ax/ret for 5 bytes.

Z80Golf, 10 bytes
00000000: 080a 0b0e 767e 23e5 ee3a                 ....v~#..:

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Disassembly
start:
  ex af,af'  ; db $3a ^ '2' ($08)
  ld a,(bc)  ; db $3a ^ '0' ($0a)
  dec bc     ; db $3a ^ '1' ($0b)
  ld c,$76   ; db $3a ^ '4' ($0e)
             ; halt         ($76)
  ld a,(hl)
  inc hl
  push hl
  xor $3a

Essentially the Hello World trick in Z80Golf.
In short, hl serves two purposes: data address for a and return address for putchar (which is at $8000 and accessed by going all the way through zeroed memory, instead of call $8000).
The instructions in the data section have to be effective no-ops (not touching hl and sp should suffice), and the last one should be 2-byte in order to shadow $76 (halt). I carefully selected the xor value so that the resulting binary has no digits ($30 to $39).
Z80Golf, 10 bytes, 2018 edition
00000000: 1416 171e 767e 23e5 ee26                 ....v~#..&

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Disassembly
start:
  inc d      ; db $26 ^ '2' ($14)
  ld d,$17   ; db $26 ^ '0' ($16)
             ; db $26 ^ '1' ($17) ; rla
  ld e,$76   ; db $26 ^ '8' ($1e)
             ; halt         ($76)
  ld a,(hl)
  inc hl
  push hl
  xor $26

Same principle, but this time we can't use ld c,$76 since it gives $36 for the xor value. So I moved to e which starts with 1e. Also, different starting address gives two different interpretations of the data section:

$16 $17: ld d,$17
$17: rla (Rotate the register A to the left)

So I had to check both are effective no-ops in this program.

Lua 5.3.1, 31 bytes
It's an old question, but I may as well join in on the fun!
For those not familiar with Lua, the # operator returns the length of an array or string, and the .. operator is for string and number concatenation. The // operator is integer divide, and is used so that there is no trailing decimal in the numbers caused by Lua treating the result of regular divisions as a float.
2014 in 31 bytes: t=#"aa"print(t..t-t..t//t..t*t) = print(2..2-2..2/2..2*2)
2015 in 36 bytes: t=#"aa"print(t..t-t..t//t..t*t+t//t) = print(2..2-2..2//2..2*2+2/2)
2016 in 33 bytes: t=#"aa"print(t..t-t..t//t..t*t+t) = print(2..2-2..2//2..2*2+2)
2017 in 38 bytes: t=#"aa"print(t..t-t..t//t..t*t+t+t//t) = print(2..2-2..2//2..2*2+2+2/2)
2018 in 33 bytes: t=#"aa"print(t..t-t..t//t..t*t*t) = print(2..2-2..2//2..2*2*2)
The solutions for other years are pretty similar. Here's a function that can do any number:
function f(y)
    t=#"a"
    s=""
    for i=t, #(y.."") do
        n=t-t
        for j=t,(y..""):sub(i,i) do
            n=n+t
        end
        s=s..n
    end
    return s
end

And here's the golfed version of the function at 108 bytes:
function f(y)t=#"a"s=""for i=t,#(y.."")do n=t-t for j=t,(y..""):sub(i,i)do n=n+t end s=s..n end return s end

Ahead, 5 bytes
ezpz
'ߞO@

' push next cell to stack
ߞ U+7DE (2014 decimal)
O output as number
@ die

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;#  -  203
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#

This language is invalid according to the community consensus on what counts as a programming language

@MilkyWay90 Um, this is from just under a year ago... I don't think there was a consensus back then. Either way, it's not like i even got an upvote, so let's just leave it, eh?

The consensus was made in 2014. You can take it up with Meta, but until then this is invalid

@MilkyWay90 Answers are not required to be in a programming language

K (ngn/k), 10 bytes
"&"*-/"U "

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and 2018 just because this is so late
K (ngn/k), 11 bytes
-"^"-*/"@!"

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EDIT: I'm unsure when this language was created, it's a variant of k which has existed for a long time, but the gitlab link seems to indicate that it may be only a year or so old (judging by commit info), so this answer may be invalid as a result

The non-competing rule has been abolished

Aheui (esotope), 8 chars (24 bytes)
반밝따바뱟해망어

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It's too easy for Aheui since it never use any kind of digits at all.
Explanation:
# An Aheui code starts with default stack "아".
반: push 2, move cursor right by 1(→).
밝: push 7, →
따: mul(push 14), →
바: push 0, →
뱟: push 2, move cursor right by 2(→→).
해: end.
망: print as integer, →
어: move cursor left by 1(←).

Note: Print instruction moves cursor in reverse direction if current storage is empty.

TI-BASIC, 7 bytes
int(₁₀^(³√(tan(cosh(cos(π

A significant improvement from the previously winning 12-byte TI-BASIC answer. Using one-byte functions allows the greatest number of chances to achieve a certain value within a given byte count.

HaykamScript, 11 bytes
s[⛉,☄]i

Explanation
s       - string
  [     - executed parameters
    ⛉   - constant for 201
    ,
    ☄   - constant for 4
  ]
i       - convert to integer

Other Years
It's pretty simple to make any other year with the same amount of bytes, Unicode codepoints 0x2600 to 0x26FF represent integers 0 to 255 in my language.
For example, to make the year this was posted:
s[⛉,☈]i

Forth, 29 bytes
hex fbc 'B' '!' / / decimal .

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*nix shell (POSIX/bourne) 41 bytes
a=.. c=. d=....;echo ${#a}${#b}${#c}${#d}
Other solutions
bashprintf %x \'—               # 13 characters Try it online!
bashecho $[x=++y+y]$?$y$[x+x]   # 25 characters Try it online!

C# .NET 75 bytes
public class p{public static void Main(){System.Console.Write((int)'ߞ');}}

Just displaying the integer value of the unicode character ߞ

Try online (also has 2015-2019)

k4, 20 bytes
*/(-)."j"$("Ud";"/")

      "j"$("Ud";"/") / convert "Ud" and "/" to underlying integer representation -> (85 100;47)
  (-).               / subtract 47 from 85 and 100 -> 38 53
*/                   / multiply over

MAWP 2.1, 15 bytes
!+!:!!-:!!$:!+:

Try it online lol.

Neim,  4  2 bytes
χρ

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Push 20, push 14 and print the entire stack

BRASCA, 4 bytes
'ߞn

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Explanation
'ߞ   - Push the character ߞ (Codepoint 2014)
  n  - Print as number

Alternative solution, 9 bytes
ll+'.E-Sn

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Explanation
ll+         - 10+10 = 20
   '.E-     - code point 46-32  = 14
       S    - Concatenate 20 and 14
        n   - Print as number

According to Wikipedia, 1 is a number. I don't know how accurate it is, though.

I'm too used to those dang numbers.

It's now fixed. No more stray numbers.

Desmoslang Assembly, 15 Bytes
PDPP+PDDP+P+POT

Swift, 92 bytes (for 2014)
func f(){print("..::.:....".components(separatedBy:":").map({x in "\(x.count)"}).joined())}

Count the number of items in the substrings between the colons, convert to strings and join them.

Commodore BASIC V2, tested on the Commodore C64, 37 PETSCII characters. Non-competing
Note that the {CRSR LEFT} in the following "source code" is the cursor left control character.
A=.↑.:A%=π+A:?A♥("╮")"{CRSR LEFT}"R╮(ST▂(A%),A)

Non-obfuscated, this would look like this:
A=.↑.:A%=π+A:PRINTASC("╮")"{CRSR LEFT}"RIGHT$(STR$(A%),A)

or on a Commodore C64 screen, it would look like this:

What’s non-competing about this? Looks fine to me

So upvote it then? idk...

I didn’t upvote because I was (and still am) confused what you meant by non-competing, if that means the answer is invalid then why should I upvote it

It is not invalid, it is non-competing or just for fun/just for a laugh.

I guess I don’t get the joke

Rattle, 1211 bytes
d|n*+b=+**b

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Explanation
d|            take "d" as input
  n           get the ASCII int value of 'd' (100)
   *          multiply (by 2 without an argument) to get 200
    +         increment by 1 to get 201
     b        add "201" to print buffer
      =+**    reset, increment, multiply by 2, multiply by 2 to get 4
          b   add "4" to print buffer, output implicitly resulting in "2014"

awk
mawk 'BEGIN { print (_+=++_)!_!!_ _^_ }'

2014

Why do you obstinately refuse to follow our established standards of formatting solutions? How many bytes is this? How do I test it?

This doesn't seem to work on Try it Online. Am I doing something wrong?

TIO link (after removing usless whitespace), byte count seems to be 29,,,

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

authors: مبین چاپ - خرید تابلو م

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/