text:
An abundance of integers!
An Abundant number is any number where the sum of its proper divisors is greater than the original number. For example, the proper divisors of 12 are:
1, 2, 3, 4, 6

And summing these results in 16. Since 16 is larger than 12, 12 is abundant. Note that this does not include "Perfect numbers", e.g. numbers that are equal to the sum of its proper divisors, such as 6 and 28.
Your task for today is to write a program or function that determines if a number is abundant or not. Your program should take a single integer as input, and output a truthy/falsy value depending on whether it is abundant or not. You can assume that the input will always be valid and greater than 0, so for bad inputs, undefined behavior is fine.
You may take your input and output in any reasonable format, for example STDIN/STDOUT, files, or arguments/return values would all be acceptable.
For reference, here are the abundant numbers up to 100:
12,
18,
20,
24,
30,
36,
40,
42,
48,
54,
56,
60,
66,
70,
72,
78,
80,
84,
88,
90,
96,
100

And more can be found on A005101
Since this is code-golf, standard loopholes are denied, and try to write the shortest possible code you can in whichever language you happen to choose!

"the first odd abundant is 945 = 3^357, the 232nd abundant number!"

The asymptotic density of abundant numbers is somewhere within the interval [0.24761748, 0.24764422]. Calculated using the source code included in this paper.

I am trying to do this in Geometry Dash. It's a nightmare

Regex (ECMAScript), 1085 855 597 536 511 508 504 bytes
Matching abundant numbers in ECMAScript regex is an entirely different beast than doing so in practically any other regex flavor. The lack of forward/nested backreferences or recursion means that it is impossible to directly count or keep a running total of anything. The lack of lookbehind makes it often a challenge even to have enough space to work in.
Many problems must be approached from an entirely different perspective, and you'll be wondering if they're solvable at all. It forces you to cast a much wider net in finding which mathematical properties of the numbers you're working with might be able to be used to make a particular problem solvable.
Back in March-April 2014 I constructed a solution to this problem in ECMAScript regex. At first I had every reason to suspect the problem was completely impossible, but then the mathematician teukon sketched an idea that made an encouraging case for making it look solvable after all – but he made it clear he had no intention of constructing the regex (he had competed/cooperated with my on constructing/golfing previous regexes, but reached his limit by this point and was content to restrict his further contributions to theorizing).
Solving unary mathematical problems in ECMAScript regex has been a fascinating journey for me, so in the past I put spoiler warnings in this post and others like it – in case others, especially those with an interest in number theory, wanted to take a crack at independently solving things like generalized multiplication. I am now removing the spoiler warnings though; they did not prove to be constructive.
Before posting my ECMAScript regex, I thought it would be interesting to analyze Martin Ender's .NET pure regex solution, ^(?!(1(?<=(?=(?(\3+$)((?>\2?)\3)))^(1+)))*1$). It turns out to be very straightforward to understand that regex, and it is elegant in its simplicity. To demonstrate the contrast between our solutions, here is a commented and pretty-printed (but unmodified) version of his regex:
# For the purpose of these comments, the input number will be referred to as N.

^(?!                  # Attempt to add up all the divisors. Since this is a regex and we
                      # can only work within the available space of the input, that means
                      # if the sum of the divisors is greater than N, the attempt to add
                      # all the divisors will fail at some point, causing this negative
                      # lookahead to succeed, showing that N is an abundant number.

  (1                  # Cycle through all values of tail that are less than N, testing
                      # each one to see if it is a divisor of N.

    (?<=              # Temporarily go back to the start so we can directly operate both
                      # on N and the potential divisor. This requires variable-length
                      # lookbehind, a .NET feature – even though this special case of
                      # going back to the start, if done left-to-right, would actually be
                      # very easy to implement even in a regex flavour that has no
                      # lookbehind to begin with. But .NET evaluates lookbehinds right
                      # to left, so please read these comments in the order indicated,
                      # from [Step 1] to [Step 7]. The comment applying to entering the
                      # lookahead group, [Step 2], is shown on its closing parenthesis.
      (?=             # [Step 3] Since we're now in a lookahead, evaluation is left to
                      #          right.
        (?(\3+$)      # [Step 4] If \3 is a divisor of N, then...
          (           # [Step 5] Add it to \2, the running total sum of divisors:
                      #          \2 = \2 + \3
            (?>\2?)   # [Step 6] Since \2 is a nested backref, it will fail to match on
                      #          the first iteration. The "?" accounts for this, making
                      #          it add zero to itself on the first iteration. This must
                      #          be done before adding \3, to ensure there is enough room
                      #          for the "?" not to cause the match to become zero-length
                      #          even if \2 has a value.
            \3        # [Step 7] Iff we run out of space here, i.e. iff the sum would
                      #          exceed N at this point, the match will fail, making the
                      #          negative lookahead succeed, showing that we have an
                      #          abundant number.
          )

        )
      )               # [Step 2] Enter a lookahead that is anchored to the start due to
                      #          having a "^" immediately to its right. The regex would
                      #          still work if the "^" were moved to the left of the
                      #          lookahead, but would be slightly slower, because the
                      #          engine would do some spurious matching before hitting
                      #          the "^" and backtracking.
      ^(1+)           # [Step 1] \3 = number to test for being a potential divisor – its
                      #               right-side-end is at the point where the lookbehind
                      #               started, and thus \3 cycles through all values from
                      #               1 to N-1.
    )
  )*1$                # Exclude N itself from being considered as a potential divisor,
                      # because if we included it, the test for proper abundance would be
                      # the sum of divisors exceeding 2*N. We don't have enough space for
                      # that, so instead what would happen if we did not exclude N as a
                      # divisor would be testing for "half-abundance", i.e. the sum of
                      # all divisors other than N exceeding N/2. By excluding N as a
                      # divisor we can let our threshold for abundance be the sum of
                      # divisors exceeding N.
)

Try the .NET regex online
Now, back to my ECMAScript regex. First, here it is in raw, whitespace-and-comment-free format:
^(?=(((?=(xx+?)\3+$)(x+)\4*(?=\4$))+(?!\3+$)(?=(xx(x*?))\5*$)x)(x+))(?=\1(x(x*))(?=\8*$)\6\9+$)(?=(.*)((?=\8*$)\5\9+$))(?=(x*?)(?=(x\11)+$)(?=\12\10|(x))(x(x*))(?=\15*$)(?=\11+$)\11\16+$)(?=(x(x*))(?=\17*$)\7\18+$)((?=(x*?(?=\17+$)(?=\17+?(?=((xx(x*))(?=\18+$)\22*$))(x+).*(?=\17$)\24*(?=\24$)(?!(xx+)\25*(?!\22+$)\25$)\22+$)((?=(x\7)+$)\15{2}\14|)))(?=.*(?=\24)x(x(x*))(?=\28*$)\23\29*$)(?=.*(x((?=\28*$)\22\29+$)))(.*(?!\30)\20|(?=.*?(?!x\20)(?=\30*$)(x(x*))(?=\33*$)(?=\31+$)\31\34+$).*(?=\33\21$)))+$
(change \14 to \14? for compatibility with PCRE, .NET, and practically every other regex flavour that's not ECMAScript)
Try it online!
Try it online! (faster, 537 byte version of the regex)
And now a brief summary of the story behind it.
At first it was very non-obvious, to me at least, that it was even possible to match primes in the general case. And after solving that, the same applied to powers of 2. And then powers of composite numbers. And then perfect squares. And even after solving that, doing generalized multiplication seemed impossible at first.
In an ECMAScript loop, you can only keep track of one changing number; that number cannot exceed the input, and has to decrease at every step. My first working regex for matching correct multiplication statements A*B=C was 913 bytes, and worked by factoring A, B, and C into their prime powers – for each prime factor, repeatedly divide the pair of prime power factors of A and C by their prime base until the one corresponding to A reaches 1; the one corresponding to C is then compared to the prime power factor of B. These two powers of the same prime were "multiplexed" into a single number by adding them together; this would always be unambiguously separable on each subsequent iteration of the loop, for the same reason that positional numeral systems work.
We got multiplication down to 50 bytes using a completely different algorithm (which teukon and I were able to arrive at independently, though it took him only a few hours and he went straight to it, whereas it took me a couple days even after it was brought to my attention that a short method existed): for A≥B, A*B=C if and only if C is the smallest number which satisfies C≡0 mod A and C≡B mod A-1. (Conveniently, the exception of A=1 needs no special handling in regex, where 0%0=0 yields a match.) I just can't get over how neat it is that such an elegant way of doing multiplication exists in such a minimal regex flavour. (And the requirement of A≥B can be replaced with a requirement that A and B are prime powers of the same power. For the case of A≥B, this can be proven using the Chinese remainder theorem.) See this post for a more in-depth explanation.
If it had turned out that there was no simpler algorithm for multiplication, the abundant number regex would probably be on the order of ten thousand bytes or so (even taking into account that I golfed the 913 byte algorithm down to 651 bytes). It does lots of multiplication and division, and ECMAScript regex has no subroutines.
I started working on the abundant number problem tangentially in 23 March 2014, by constructing a solution for what seemed at the time to be a sub-problem of this: Identifying the prime factor of highest multiplicity, so that it could be divided out of N at the start, leaving room to do some necessary calculations. At the time this seemed to be a promising route to take. (My initial solution ended up being quite large at 326 bytes, later golfed down to 185 bytes.) But the rest of the method teukon sketched would have been extremely complicated, so as it turned out, I took a rather different route. It proved to be sufficient to divide out the largest prime power factor of N corresponding to the largest prime factor on N; doing this for the prime of highest multiplicity would have added needless complexity and length to the regex.
What remained was treating the sum of divisors as a product of sums instead of a straight sum. As explained by teukon on 14 March 2014:

We're given a number n = p0a0p1a1...pk-1ak-1.  We want to handle the sum of the factors of n, which is (1 + p0 + p02 + ... + p0a0)(1 + p1 + p12 + ... + p1a1)...(1 + pk-1 + pk-12 + ... + pk-1ak-1).

It blew my mind to see this. I had never thought of factoring the aliquot sum in that way, and it was this formula more than anything else that made the solvability of abundant number matching in ECMAScript regex look plausible.
In the end, instead of testing for a result of addition or multiplication exceeding N, or testing that such a result pre-divided by M exceeds N/M, I went with testing if a result of division is less than 1. I arrived at the first working version on 7 April 2014.
The full history of my golf optimizations of this regex is on github. At a certain point one optimization ended up making the regex much slower, so from that point on I maintained two versions. They are:
regex for matching abundant numbers.txt
regex for matching abundant numbers - shortest.txt
These regexes are fully compatible with both ECMAScript and PCRE, but a recent optimization involved using a potentially non-participating capture group \14, so by dropping PCRE compatibility and changing \14? to \14 they can both be reduced by 1 byte.
Here is the smallest version, with that optimization applied (making it ECMAScript-only), reformatted to fit in a StackExchange code block with (mostly) no horizontal scrolling needed:
# Match abundant numbers in the domain ^x*$ using only the ECMAScript subset of regex
# functionality. For the purposes of these comments, the input number = N.
^
# Capture the largest prime factor of N, and the largest power of that factor that is
# also a factor of N. Note that the algorithm used will fail if N itself is a prime
# power, but that's fine, because prime powers are never abundant.
(?=
  (                      # \1 = tool to make tail = Z-1
    (                    # Repeatedly divide current number by its smallest factor
      (?=(xx+?)\3+$)
      (x+)\4*(?=\4$)
    )+                   # A "+" is intentionally used instead of a "*", to fail if N
                         #  is prime. This saves the rest of the regex from having to
                         #  do needless work, because prime numbers are never abundant.
    (?!\3+$)             # Require that the last factor divided out is a different prime.
    (?=(xx(x*?))\5*$)    # \5 = the largest prime factor of N; \6 = \5-2
    x                    # An extra 1 so that the tool \1 can make tail = Z-1 instead of just Z
  )
  (x+)                   # Z = the largest power of \5 that is a factor of N; \7 = Z-1
)
# We want to capture Z + Z/\5 + Z/\5^2 + ... + \5^2 + \5 + 1 = (Z * \5 - 1) / (\5 - 1),
# but in case Z * \5 > N we need to calculate it as (Z - 1) / (\5 - 1) * \5 + 1.
# The following division will fail if Z == N, but that's fine, because no prime power is
# abundant.
(?=
  \1                     # tail = (Z - 1)
  (x(x*))                # \8   = (Z - 1) / (\5 - 1); \9 = \8-1
  # It is guaranteed that either \8 > \5-1 or \8 == 1, which allows the following
  # division-by-multiplication to work.
  (?=\8*$)
  \6\9+$
)
(?=
  (.*)                   # \10 = tool to compare against \11
  (                      # \11 = \8 * \5  =  (Z - 1) / (\5 - 1) * \5; later, \13 = \11+1
    (?=\8*$)
    \5\9+$
  )
)
# Calculate Q = \15{2} + Q_R = floor(2 * N / \13). Since we don't have space for 2*N, we
# need to calculate N / \13 first, including the fractional part (i.e. the remainder),
# and then multiply the result, including the fractional part, by 2.
(?=
  (x*?)(?=(x\11)+$)      # \12 = N % \13; \13 = \11 + 1
  (?=\12\10|(x))         # \14 = Q_R = floor(\12 * 2 / \13)
                         #     = +1 carry if \12 * 2 > \11, or NPCG otherwise
  (x(x*))                # \15 = N / \13; \16 = \15-1
  (?=\15*$)
  (?=\11+$)              # must match if \15 <  \13; otherwise doesn't matter
  \11\16+$               # must match if \15 >= \13; otherwise doesn't matter
)
# Calculate \17 = N / Z. The division by Z can be done quite simply, because the divisor
# is a prime power.
(?=
  (x(x*))                # \17 = N / Z; \18 = \17-1
  (?=\17*$)
  \7\18+$
)
# Seed a loop which will start with Q and divide it by (P^(K+1)-1)/(P-1) for every P^K
# that is a factor of \17. The state is encoded as \17 * P + R, where the initial value
# of R is Q, and P is the last prime factor of N to have been already processed.
#
# However, since the initial R would be larger than \17 (and for that matter there would
# be no room for any nonzero R since with the initial value of P, it is possible for
# \17 * P to equal N), treat it as a special case, and let the initial value of R be 0,
# signalling the first iteration to pretend R=Q. This way we can avoid having to divide Q
# and \17 again outside the loop.
#
# While we're at it, there's really no reason to do anything to seed this loop. To seed
# it with an initial value of P=\5, we'd have to do some multiplication. If we don't do
# anything to seed it, it will decode P=Z. That is wrong, but harmless, since the next
# lower prime that \17 is divisible by will still be the same, as \5 cannot be a factor
# of \17.

# Start the loop.
(
  (?=
    (                    # \20 = actual value of R
      x*?(?=\17+$)       # move forward by directly decoded value of R, which can be zero
      # The division by \17 can be done quite simply, because it is known that
      # the quotient is prime.
      (?=
        \17+?            # tail = \17 * (a prime which divides into \17)
        (?=
          (              # \21 = encoded value for next loop iteration
            (xx(x*))     # \22 = decoded value of next smaller P; \23 = (\22-1)-1
            (?=\18+$)    # iff \22 > \17, this can have a false positive, but never a false negative
            \22*$        # iff \22 < \17, this can have a false positive, but never a false negative
          )
        )
        # Find the largest power of \22 that is a factor of \17, while also asserting
        # that \22 is prime.
        (x+)             # \24 = the largest power of \22 that is a factor of \17
        .*(?=\17$)
        \24*(?=\24$)
        (?!
          (xx+)\25*
          (?!\22+$)
          \25$
        )
        \22+$
      )
      (
        (?=(x\7)+$)      # True iff this is the first iteration of the loop.
        \15{2}\14        # Potentially unset capture, and thus dependent on ECMAScript
                         # behavior. Change "\14" to "\14?" for compatibility with non-
                         # ECMAScript engines, so that it will act as an empty capture
                         # with engines in which unset backrefs always fail to match.
      |
      )
    )
  )
  # Calculate \30 = (\24 - 1) / (\22 - 1) * \22 + 1
  (?=
    .*(?=\24)x           # tail = \24 - 1
    (x(x*))              # \28 = (\24 - 1) / (\22 - 1); \29 = \28-1
    (?=\28*$)
    \23\29*$
  )
  (?=
    .*(x(                # \30 = 1 + \28 * \22 = (\28 - 1) / (\22 - 1) * \22 + 1; \31 = \30-1
      (?=\28*$)
      \22\29+$
    ))
  )
  # Calculate \33 = floor(\20 / \30)
  (
    .*(?!\30)\20         # if dividing \20 / \30 would result in a number less than 1,
                         # then N is abundant and we can exit the loop successfully
  |
    (?=
      .*?(?!x\20)(?=\30*$)
      (x(x*))            # \33 = \20 / \30; \34 = \33-1
      (?=\33*$)
      (?=\31+$)          # must match if \33 <  \30; otherwise doesn't matter
      \31\34+$           # must match if \33 >= \30; otherwise doesn't matter
    )
    # Encode the state for the next iteration of the loop, as \17 * \22 + \33
    .*(?=\33\21$)
  )
)+$

Comments are not for extended discussion; this conversation has been moved to chat.

-98 bytes

Python 2, 41 40 bytes
n=k=j=input()
while~k<0:j-=1;k-=j>>n%j*n

Output is via exit code, so 0 is truthy and 1 is falsy.
Try it online!
How it works
After setting all of n, k, and j to the input from STDIN, we enter the while loop. Said loop will break as soon as -k - 1 = ~k ≥ 0, i.e., k ≤ -1 / k < 0.
In each iteration, we first decrement j to consider only proper divisors of n. If j is a divisor of n, n%j yields 0 and j >> n%j*n = j/20 = j gets subtracted from k. However, if j does not divide n, n%j is positive, so n%j*n is at least n > log2 j and j >> n%j*n = j / 2n%j*n = 0 is subtracted from k.
For abundant numbers, k will reach a negative value before or when j becomes 1, since the sum of n's proper divisors is strictly greater than n. In this case, we break out of the while loop and the program finishes normally.
However, if n is not abundant, j eventually reaches 0. In this case, n%j throws a ZeroDivisionError and the program exits with an error.

~k<0 is fancy, but I think -1<k also does the trick ;)

Brachylog, 5 bytes
fk+>?

Try it online!
Explanation
f           Factors
 k          Knife: remove the last one (the input itself)
  +         Sum
   >?       Stricly greater than the Input

Mathematica, 17 bytes
Tr@Divisors@#>2#&

Explanation
Tr@                 The sum of the main diagonal of
   Divisors@         the list of divisors of
            #         the first argument
             >      is greater than
              2#      twice the first argument.
                &   End of function.

I'm surprised Mathematica has no built in for this..

@MrPaulch Considering the length of the program though, the builtin may very well be longer in name >.>

@ConorO'Brien If it existed, it would probably be AbundantNumberQ, so it would save a couple bytes :)

Jelly, 3 bytes
Æṣ>

Try it online!
How it works
Æṣ>  Main link. Argument: n

Æs   Get the proper divisor sum of n.
  >  Test if the result is greater than n.

Python, 44 bytes
lambda n:sum(i*(n%i<1)for i in range(1,n))>n

Try it online!

It's a shame you can't do range(n). That pesky modulus by zero

05AB1E, 5 4 bytes
-1 bytes thanks to scottinet
Ñ¨O‹

Try it online! or Try 0 to 100

Retina, 50 45 bytes
^(?!(1(?<=(?=(?(\3+$)((?>\2?)\3)))^(1+)))*1$)

Input in unary, output 1 for abundant numbers, 0 otherwise.
There is nothing Retina-specific about this solution. The above is a pure .NET regex which matches only abundant numbers.
Try it online! (Test suite that filters decimal input with the above regex.)

Retina, 34 bytes
Byte count assumes ISO 8859-1 encoding.
M!&`(1+)$(?<=^\1+)
1>`¶

^(1+)¶1\1

Input in unary, output 1 for abundant numbers, 0 otherwise.
Try it online!
Explanation
M!&`(1+)$(?<=^\1+)

We start by getting all divisors of the input. To do this, we return (!) all overlapping (&) matches (M) of the regex (1+)$(?<=^\1+). The regex matches some suffix of the input, provided that the entire input is a multiple of that suffix (which we ensure by trying to reach the beginning fo the string using only copies of the suffix). Due to the way the regex engine looks for matches, this will result a list of divisors in descending order (separated by linefeeds).
1>`¶

The stage itself simply matches linefeeds (¶) and removes them. However, the 1> is a limit, which skips the first match. So this effectively adds together all divisors except the input itself. We end up with the input on the first line and the sum of all proper divisors on the second line.
^(1+)¶1\1

Finally, we try to match the at least one more 1 on the second line than we have on the first line. If that's the case, the sum of proper divisors exceeds the input. Retina counts the number of matches of this regex, which will be 1 for abundant numbers and 0 otherwise.

It's always amazed me how you can do math in retina. I'd love to see an explanation! :)

@DJMcMayhem Sorry forgot to add that earlier. Done.

x86 machine code, 25 bytes
00000000: 8bc8 d1e9 33db 33d2 50f7 f158 85d2 7502  ....3.3.P..X..u.
00000010: 03d9 7204 e2f0 3bc3 c3                   ..r...;..

Listing:
8B C8       MOV  CX, AX         ; counter starts at N / 2
D1 E9       SHR  CX, 1          ; divide by 2
33 DB       XOR  BX, BX         ; clear BX (running sum of factors)
    DIV_LOOP:
33 D2       XOR  DX, DX         ; clear DX (high word for dividend)
50          PUSH AX             ; save original dividend
F7 F1       DIV  CX             ; DX = DX:AX MOD CX, AX = DX:AX / CX
58          POP  AX             ; restore dividend (AX was changed by DIV)
85 D2       TEST DX, DX         ; if remainder (DX) = 0, CX is a divisor
75 02       JNZ  CONT_LOOP      ; if not, continue loop to next
03 D9       ADD  BX, CX         ; add number to sum
    CONT_LOOP:
72 04       JC   END_ABUND      ; if CF=1, BX has unsigned overflow it is abundant
E2 F0       LOOP DIV_LOOP
3B C3       CMP  AX, BX         ; BX<=AX -> CF=0 (non-abund), BX>AX -> CF=1 (abund)
    END_ABUND:
C3          RET                 ; return to caller

Callable function, input N in AX.  Output CF=1 if abundant, CF=0 if not abundant.
Try it online!
Props:

Fixed for 16-bit overflow (+5 bytes). Thanks @deadcode for the suggestions!
Simplified return logic (-3 bytes). Thx to help from @deadcode once again.
Use TEST instead of CMP (-1 byte). Thx to @l4m2, and make sure to upvote @l4m2's shorter x86 answer!
Thx to @deadcode for the TIO link

I would suggest replacing JLE with JBE to double the range of numbers this can test before overflows start causing it to give false negatives. Then instead of starting to fail at 12600 (aliquot sum 35760), it'll start to fail at 25200 (aliquot sum 74744). Even better would be to also detect the carry flag and treat that as an abundant number without needing to calculate the actual >16 bit sum.

Good catch @Deadcode. I've updated the code for jump below instead of jump less. I see what you mean, doing a JC after the ADD BX,CX will catch the unsigned overflow there and make it correct up until N=65535.  Complicates flag testing and return state a bit, since previously CF implied false.  Updated with fix also.

Nice fixes. But you didn't correctly update your comments; the reverse of JBE is JA (not JG), and these both look at only CF and ZF, ignoring SF and OF.

You can save 3 bytes by inverting the specification of your return value, with CF being set if abundant and clear if not. But I'd recommend doing the edit to correct the output documentation first, so it looks nice in the edit history.

Also, to keep things simple, the specification should be that the return value is in the carry flag, and none of that fuss with the other flags. The caller should use JC or JNC to act on whether the number is abundant or not.

Also, it is misleading to call the unassembled version "Ungolfed". If assembling it with an assembler results in the machine code you have shown, then it is golfed. The fact that it is pretty-printed and has comments has no effect on the output when the program is assembled. Calling it "Unassembled" would be correct. The only reason you should call such a listing "Ungolfed" is if you did manual optimizations directly to the machine language bytes that cannot be, or have not been, done in an assembly language.

Thanks so much for all of your help and your comments. I've updated the inline documentation and removed the term ungolfed. Honestly never gave it much thought but I see your point about it, as it isn't any different with the exception of inline comments. Also agree about makint the return flags clearer as well. Will work on that in a bit. Thanks for the attention and the assistance!

@Deadcode, once again very clever observation and suggestion about simplifying return flags.  I've implemented your change and it is now 3 bytes smaller.  Bravo!

Beautiful, nicely done!

SHR CX, 1 => DEC CX ? CMP DX, 0 => TEST DX, DX?

Thanks @l4m2!  Updated to use TEST for -1 byte.

Oops. You should use the SHR CX, 1 → DEC CX suggestion too. Although it'll make the function approximately twice as slow, it's good golf as it saves 1 byte. The only important thing is that all proper divisors still get tested and N itself is skipped.

Try it online!

@Deadcode thanks! Updated and added!

Note that the TIO is not executing exactly the same machine code, as it needs to use 0x66 and 0x67 prefixes to do 16-bit operations, but I still think it's much better than nothing. :-) And if you really want to change your answer from a macro to a function, please adjust the TIO too, if you still want to use it. (FWIW I think it's fine as a macro.) Please use a direct TIO link – I only used tinyurl because the URL was too big for a comment! I guess I need to put a TIO in the comments of @l4m2's version too. I had forgotten about it, and it does legitimately save 3 bytes.

Java 8, 53 bytes ( a lot more if you include the ceremonial code )
return IntStream.range(1,n).filter(e->n%e<1).sum()>n;

Try it online
Explanation :
IntStream.range(1,n) \\ numbers from 1 to n-1
filter(e->n%e<1)     \\ filter in numbers that perfectly divide the number n
sum()>n              \\ sum and compare to original number

Great answer, but with Java 8 you must include the function in your byte-count. Then again, you can drop the return if I'm not mistaken, so it will be even shorter: n->IntStream.range(1,n).filter(e->n%e<1).sum()>n (not 100% if this is correct, I almost never program in Java 8). Welcome to PPCG!

The correct count via standard count would be n->java.util.stream.IntStream.range(1,n).filter(e->n%e<1).sum()>n for 65 bytes (assuming I got the package right off the top of my head)

Actually, 5 bytes
;÷Σ½>

Try it online!
;     # Duplicate input
 ÷    # Get divisors
  Σ   # Sum
   ½  # Divide by 2 (float)
    > # Test is greater than input

05AB1E, 4 bytes
Ñ¨O‹

Try it online!
How it works
Ñ        #list of divisors
 ¨       #remove last element (i.e the input from the list of factors)
  O      #sum the list
   ‹     #is this sum less than the input?

Sorry to post in old question, I was just going through old posts for practise and noticed my solution was shorter than the next best 05AB1E solution.

Sorry to post in old question Don't worry about it! I'm always happy to see answers on my old challenges, and it's actually encouraged around here. :)

Regex (Perl/PCRE+(?^=)RME), 36 bytes
(?=((\2x+?|^x)(?^=\2+$))+)(\2x+)\3+$

Try it on replit.com (RegexMathEngine)
Takes its input in unary, as the length of a string of xs.
This is based on my 43 byte .NET version; for the full explanation, see below. Lookinto, a new feature in RegexMathEngine, allows (?<=(?=\2+$)^.*) to be shortened to the equivalent (?^=\2+$).
This is not just the shortest, but the fastest under RegexMathEngine, even though the 43 byte .NET regex it's based on is the slowest under .NET (beaten by both 44 byte regexes in speed).
Regex (Java / .NET) / Retina, 43 bytes
(?=((\2x+?|^x)(?<=(?=\2+$)^.*))+)(\2x+)\3+$

Try it online! - Java (very slow)
Try it online! - Regex (.NET)
Try it online! - Retina
This is 2 bytes shorter than Martin Ender's answer. Its structure is quite different, and it's more portable to different regex engines.
It works by first adding up as many proper divisors of N as it can, in strict order from smallest to largest, while having a sum not greater than N. Then it checks if there exists a larger proper divisor than the last one added to the sum; if there does, N is an abundant number.
# For the purpose of these comments, the input number will be referred to as N.

                      # No anchor needed, because \2 cannot be captured if
                      # matching is started anywhere besides the beginning.
(?=
    (                 # Add up all values of \2 until there is no more room in
                      # N to keep going
        (\2x+?|^x)    # \2 = the smallest value greater than the previous value
                      # of \2 which satisfies the following condition; if we're
                      # at the beginning, \2 = 1
        (?<=          # Lookbehind - evaluated from right to left
            (?=\2+$)  # Assert that N is divisible by \2
            ^.*       # Jump to the beginning of the input string, then execute
                      # the above lookahead
        )
    )+
)
(\2x+)                # Match if there exists another proper divisor of N, \3,
\3+$                  # such that \3 is greater than the last value of \2. If
                      # there does, that means there was insufficient room in N
                      # to add up all of its proper divisors, and N is abundant.

Regex (Pythonregex / .NET) / Retina, 47 bytes
(?=((\3x+?|^x)(?<=(?=\2+$)^.*(\2)))+)(\2x+)\4+$

Try it online! - Python import regex
Try it online! - .NET
Try it online! - Retina
This is a port of the 43 byte Java/.NET version, avoiding a nested backreference in order to add Pythonregex support. The way in which this is done (placing the copy inside the lookbehind) removes Java support.
Regex (Perl / PCRE2 v10.35+), 49 bytes
This is a direct port of the .NET regex, emulating variable-length lookbehind using a recursive subroutine call.
(?=((\2x+?|^x)((?<=(?=^\2+$|(?3)).)))+)(\2x+)\4+$

Try it online! - Perl (slow)
Attempt This Online! - PCRE2 v10.40+
(?=
    (
        (\2x+?|^x)
        ((?<=          # (?3) calls this
            (?=        # Circumvent the constant-width limitation of lookbehinds
                       # in PCRE by using a lookahead inside the lookbehind
                ^\2+$  # This is the payload of the emulated variable-length
                       # lookbehind, same as the one in the .NET regex
            |
                (?3)   # Recursive call - this is the only alternative that can
                       # match until we reach the beginning of the string
            )
            .          # Go back one character at a time, trying the above
                       # lookahead for a match each time
        ))
    )+
)
(\2x+)
\4+$

Regex (Java / Pythonregex / .NET) / Retina, 50 bytes
(?=((?=(\3|^))(\2x+?)(?<=(?=\3+$)^.*))+)(\3x+)\4+$

Try it online! - Java (very slow)
Try it online! - Python import regex
Try it online! - .NET
This is a port of the 43 byte Java/.NET version, avoiding a nested backreference in order to add Pythonregex support, while keeping Java support.
Regex (Perl / PCRE), 58 bytes
Porting the PCRE2 regex to PCRE1 requires working around some bugs:
(?=((?=(\3|^))(\2x+?)((?<=(?=z|^\3+$|(?4)).)))+)(\3x+)\5+$

The (\2x+?|^x) had to be changed to (?=(\3|^))(\2x+?), copying \3 into \2 and \2 into \3 (emulating a nested backref using a forward backref), because using a backreference inside its own group forces that group to be atomic in PCRE1, and in PCRE2 up to v10.34. And a fake always-false alternative z had to be added to the recursive subroutine to prevent the error message "recursive call could loop indefinitely" in PCRE1.
Try it online! - Perl (slow)
Try it online! - PCRE1
Try it online! - PCRE2 v10.33
Attempt This Online! - PCRE2 v10.40+

Regex (.NET) / Retina, 44 bytes
^(?!((?<=(?=(?(\3+$)((?>\2?)\3)))(^x*))x)*$)

Try it online! - Regex (.NET)
Try it online! - Retina
^(?!(x(?=(x*$)(?<=(?(^\2+)(\2(?>\3?))))))*$)

Try it online! - Regex (.NET)
Try it online! - Retina
This is based on Martin Ender's answer, but saving 1 byte by testing zero as a potential divisor of \$n\$ (it never will be, so this is harmless), thus skipping the need to explicitly exclude \$n\$ itself as a divisor. Another upshot of this is that we can extend the valid domain from ^\x+$ to ^\x*$, giving the correct answer (non-match, i.e. false) for an input of zero.
I have included versions with both smallest-to-largest and largest-to-smallest divisor summing order. The length for both is 44 bytes. Only the largest-to-smallest version is commented below, because it's simpler to explain, and I already commented the original 45 byte smallest-to-largest version in my ECMAScript answer.
^(?!                  # Attempt to add up all the divisors. Since this is a regex and we
                      # can only work within the available space of the input, that means
                      # if the sum of the divisors is greater than N, the attempt to add
                      # all the divisors will fail at some point, causing this negative
                      # lookahead to succeed, showing that N is an abundant number.
  (x                  # Cycle through all positive values of tail that are less than N,
                      # testing each one to see if it is a divisor of N. Start at N-1.
    (?=               # Do the below operations in a lookahead, so that upon popping back
                      # out of it our position will remaing the same as it is here.
      (x*$)           # \2 = tail, a potential divisor; go to end to that the following
                      # lookbehind can operate on N as a whole.
      (?<=            # Switch to right-to-left evaluation so that we can operate both
                      # on N and the potential divisor \2. This requires variable-length
                      # lookbehind, a .NET feature. Please read these comments in the
                      # order indicated, from [Step 1] to [Step 4].
        (?(^\2+)      # [Step 1] If \2 is a divisor of N, then...
          (           # [Step 2] Add it to \3, the running total sum of divisors:
                      #          \3 = \3 + \2
            \2        # [Step 4] Iff we run out of space here, i.e. iff the sum would
                      #          exceed N at this point, the match will fail, making the
                      #          negative lookahead succeed, showing that we have an
                      #          abundant number.
            (?>\3?)   # [Step 3] Since \3 is a nested backref, it will fail to match on
                      #          the first iteration. The "?" accounts for this, making
                      #          it add zero to itself on the first iteration. This must
                      #          be done before adding \2, to ensure there is enough room
                      #          for the "?" not to cause the match to become zero-length
                      #          even if \3 has a value.
          )
        )
      )
    )
  )*                  # By using "*" instead of "+" here, we can avoid categorizing zero
                      # as an abundant number.
  $                   # We can only reach this point if all proper divisors of N, all the
                      # way down to \2 = 1, been successfully summed into \3, which would
                      # mean N is not an abundant number.
)

As far as speed goes, the smallest-to-largest version is faster than the largest-to-smallest version (both in .NET and their ports below). Both are faster than the 43 byte version under .NET.
Regex (Perl/PCRE+(?^=)RME), 42 bytes
^(?!(x(?=(x*))(?^=(?(?=\2+$)(\3?+\2))))*$)

Try it on replit.com (RegexMathEngine)
This is a port of the largest-to-smallest 44 byte .NET regex, using lookinto instead of variable-length lookbehind. Although the Perl/PCRE syntax for lookahead conditionals is more verbose, this is compensated for by being able to use a possessive quantifier.
Regex (Perl/PCRE+(?^=)RME), 44 bytes
^(?!((?^0=(x*))(?^=(?(?=\2+$)(\3?+\2)))x)*$)

Try it on replit.com (RegexMathEngine)
This is a port of the smallest-to-largest 44 byte .NET regex, again using lookinto instead of variable-length lookbehind. This one is less conducive to being ported, and I had to use (?^0=), lookinto accessing the in-progress current match. I've not decided how to finalize the current behavior of this, so it's likely this version will not work with a future version of lookinto.
Regex (PCRE2), 73 bytes
This is a direct port of the 44 byte .NET regex (based on Martin Ender's regex). It doesn't port well to PCRE, because it changes the value of a capture group inside the lookbehind – but upon the return of any subroutine in PCRE, all capture groups are reset to the values they had upon entering the subroutine. So the only way to port it is to change the main loop from iteration into recursion, resulting in a large and very slow regex:
^(?!(x(?=(x*$)((?<=(?=^(?(?=\2+$)(?=(\4?+\2))).*(?=\2$)(?1)|(?3)).)))|$))

Try it online! - PCRE2 v10.33 (PHP)
Try it on regex101 - PCRE2 v10.38+
Attempt This Online! - PCRE2 v10.40+ (PHP)
Attempt This Online! - PCRE2 v10.40+ (C++)

Variable-length lookbehind emulation? Whatever next?

Julia, 25 bytes
a(n,r=1:n)=(n.%r.<1)'r>2n
Attempt This Online!

Powershell, 51 49 Bytes
param($i)((1..$i|?{!($i%$_)})-join"+"|iex)-gt2*$i

I wish I could remove some brackets.
-2 thanks to AdmBorkBork, instead of not counting the input in the initial range, we just take it into account in the final check.
Loop through range of 1.. to the $input, minus 1, find where (?) the inverse modulo of input by the current number is $true (aka only 0) - then -join all of those numbers together with + and iex the resulting string to calculate it, then see if the sum of these parts is greater than the input.
PS C:\++> 1..100 | ? {.\abundance.ps1 $_}
12
18
20
24
30
36
40
42
48
54
56
60
66
70
72
78
80
84
88
90
96
100

You can save two bytes by counting the top value and checking that it's bigger than 2x the input -- param($i)((1..$i|?{!($i%$_)})-join"+"|iex)-gt2*$i

PARI/GP, 15 bytes
n->sigma(n)>2*n

The variant n->sigma(n,-1)>2 is, unfortunately, longer.

k, 19 16 15 bytes
{x<+/&~(!x)!'x}

Returns 1 for true, and 0 for false.
Try it online!
{             } /function(x)
       (!x)     /{0, 1, ..., x-1}
            '   /for each n in {0, 1, ..., x-1}:
           ! x  /    do (x mod n)
      ~         /for each, turn 0 -> 1, * -> 0 (map not)
     &          /get indices of 1's
   +/           /sum (fold add)
 x<             /check if x < the sum

QBIC, 22 bytes
:[a/2|~a%b|\p=p+b}?p>a

This is an adaptation to the QBIC primality test. Instead of counting the divisors and checking if it's less than three, this sums the proper divisors. This runs only along half of 1 to n, where the primality test runs through 1 to n completely.
Explanation:
:       Get the input number, 'a'
[a/2|   FOR(b=1, b<=(a/2), b++)
~a%b    IF a MOD b != 0  --> QBasic registers a clean division  (0) as false.
        The IF-branch ('|') therefor is empty, the code is in the ELSE branch ('\')
|\p=p+b THEN add b to runnning total p
}       Close all language constructs: IF/END IF, FOR/NEXT
?p>a    Print '-1' for abundant numbers, 0 otherwise.

Pure Bash, 37 bytes
for((;k++<$1;s+=$1%k?0:k)){((s>$1));}

Thanks to @Dennis for rearranging the code -- saving 6 bytes and eliminating the incidental output to stderr.
The input is passed as an argument.
The output is returned in the exit code: 0 for abundant, 1 for not abundant.
Output to stderr should be ignored.
Test runs:
for n in {1..100}; do if ./abundant "$n"; then echo $n; fi; done 2>/dev/null
12
18
20
24
30
36
40
42
48
54
56
60
66
70
72
78
80
84
88
90
96
100

You can save 6 bytes while avoiding stray output to STDERR. https://tio.run/nexus/bash#S04sUbBTSEwqzUtJzCtRsLFRUHf1d1P/n5ZfpKFhna2tbaNiaF2sbatiqJptb2CVralZraFRbKdiqKlpXfsfqJSLC6hUIU8hM0@h2lBPz9DAoNZaISWfizMpsTgDYbBKnoKamkJqckY@kMmVkp@X@h8A

Japt, 9 7 6 bytes
<Uâ1 x

Saved 2 bytes thanks to ETHproductions. Saved 1 byte thanks to obarakon.
Try it online!

9 characters, 10 bytes.

@Metoniem I'm sure â is 1 byte, in unicode at least (0xE2).

@Metoniem Japt uses the ISO-8859-1 encoding, in which â is a single byte.

If â is given a truthy argument, it will remove the actual number from the remaining list, so you can do â1 x >U to save a couple bytes :-)

@TomDevs Nice! You can do <Uâ1 x to save a byte. Japt adds the U in front of the program.

@ETHproductions Yeah, I only just now figured out they're not all counted in the same encoding. Oops :(

You can ditch the U to save another byte.

JavaScript (ES6), 33 bytes

let g =
x=>(f=n=>--n&&n*!(x%n)+f(n))(x)>x
<input type=number min=1 value=1 step=1 oninput="O.innerHTML=g(+value)"><br>
<pre id=O>false</pre>

I was sure a recursive answer would be best but I didn't think it would be this good.

MATL, 6 bytes
Z\sGE>

Outputs 1 for abundant numbers, 0 otherwise.
How it works
Z\      % list the divisors of the implicit input
s       % add them
G       % push the input again
E       % double it
>       % compare
        % implicitly display result

Cubix, 38 bytes
/..?%?(O;0I:^.<.>;rrw+s;rUO?-<...O0L.@

Try it here
      / . .
      ? % ?
      ( O ;
0 I : ^ . < . > ; r r w
+ s ; r U O ? - < . . .
O 0 L . @ . . . . . . .
      . . .
      . . .
      . . .

0I: - sets up the stack with 0, n, n (s, n, d)
^ - start of the loop
)? - decrement d and test for 0.  0 exits loop
%? - mod against n and test.  0 causes ;rrw+s;rU which rotates s to top and adds d, rotates s to bottom and rejoins loop
;< - Cleanup and rejoin the loop.
On exiting loop
;< - Remove d from stack and redirect
-? - Remove n from s and test, 0 LOU@ turns left, outputs and exits, negatives 0O@ push zero, output and exits.  positives ;O remove difference and outputs n. The path then goes through to the left turn which redirects to the @ exit

APL (Dyalog Unicode), 11 bytesSBCS
-2 bytes thanks to @Adám

⊢<1⊥∘⍸0=⍳|⊢

Try it online!
Explanation:
⊢<1⊥∘⍸0=⍳|⊢ ⍝ Monadic function train
        ⍳   ⍝ Generate integers from 1 to input
        |⊢ ⍝ For each of the above, modulo the input
      0=   ⍝ Return a boolean vector where ones correspond
           ⍝ to zeroes
     ⍸      ⍝ Return the indices of the ones in the above vector
           ⍝ This results in a vector of the proper divisors of our input
  1⊥∘      ⍝ Composed with the above, decode into unary base.
           ⍝ This has the same effect as summing the vector, but is used
           ⍝ here to comply with the function train format.
⊢<         ⍝ Is the above sum greater than our input?

-2 with ⊢<1⊥∘⍸0=⍳|⊢ as per this tip.

Arn, 12 bytes
Ý░™€ºa^¬"/”&

Try it!
Explained
Unpacked: <(+\$v{!%v}1->
  _                    Variable initialized to STDIN; implied
<                      Is less than
  (                    Begin expression
      \                Fold with
    +                  Addition
        $              Filtered list
          v{           Block with key of v
            !          Boolean not
                _      Implied
              %        Modulo
                v
           }           End of block
             1         Literal one
           ->          Exclusive range
             _         Implied
  )                    End of expression, implied

I should probably add a builtin fix for getting factors...

Batch, 84 bytes
@set/ak=%1*2
@for /l %%j in (1,1,%1)do @set/ak-=%%j*!(%1%%%%j)
@cmd/cset/a"%k%>>31

Outputs -1 for an abundant number, 0 otherwise. Works by subtracting all the factors from 2n and then shifting the result 31 places to extract the sign bit. Alternative formulation, also 84 bytes:
@set k=%1
@for /l %%j in (1,1,%1)do @set/ak-=%%j*!(%1%%%%j)
@if %k% lss -%1 echo 1

Outputs 1 for an abundant number. Works by subtracting all the factors from n and then comparing the result to to -n. (set/a is Batch's only way of doing arithmetic so I can't easily adjust the loop.)

"(%1%%%%j)"  oh, batch :)

Perl 6, 72 24 bytes
{$_ <sum grep $_%%*,^$_}

Program argument: a.
Generate a list from 1..a.
Take all the numbers that are divisors of a.
Sum them.
Check if that sum is greater than a.

Thanks to @b2gills.

Every occurrence of $^a after the first one can be shortened to just $a. but it is even shorter if you write it as {$_ <sum grep $_%%*,^$_} Also looking at an earlier version, [+](LIST) works (no spaces)

@BradGilbertb2gills Thanks! :)

Pyth, 11 bytes
>sPf!%QTS

Old:
L!%Qb>sPy#S

I can't use !% as a pfn for #, because it's two functions. Makes me sad :(.

L!%Qb>sPy#SQ    Program's argument: Q
L!%Qb           Define a lambda y, that takes b as an argument
 !%Qb           Return true if Q is divisible by b
          S     Make a range 1..Q
        y#      Filter that range with the lambda (y)
       P        Remove the last element (Q itself)
      s         Sum them
     >     Q    Check if that sum is greater than the program's argument

Not defining a function seems to be shorter: >sPf!%QTS

Ruby, 39 32 bytes
->n{n<(1...n).sum{|x|n%x>0?0:x}}

Try it online!
Updated to Ruby 2.6

J, 19 bytes
Thanks to Conor O'Brien for cutting it to 19 bytes!
<[:+/i.#~i.e.]%2+i.
Previous: (34 bytes)
f=:3 :'(+/((i.y)e.y%2+i.y)#i.y)>y'
Returns 1 if it's abundant and 0 if it's not.
Output:
   f 3
0
   f 12
1
   f 11
0
   f 20
1

Welcome to PPCG! We allow anonymous functions, so you can remove the leading f=: as part of your byte count. Also, you can get down to 19 by converting to a tacit verb: <[:+/i.#~i.e.]%2+i.

Thanks for the advice! However, could you please explain the cap verb ([:) and the switch verb (~). I don't really get what they're supposed to do in this tacit verb.

~ switches so it's i.e.#i. but what is the purpose of [: ?

so you know about forks, right? (f g h) y' is the same as (f y) g (h y). When fis a cap,([: g h) y` is roughly the same as `g h y`. As for `~`, this switches the left and right arguments. It is important to know that `~` is not a verb but is actually an adverb. It modifies a verb. For example, we could have something like `2 %~ 8`. Here, `~` modifies `%` to switch its arguments, so the expression is equivalent to `8 % 2`.

In the fork chain, #~ is evaluated after the verbs to its right are executed, so it's left argument becomes the result on the right

Thanks for so much for the explanation. It makes more sense now.

My pleasure! If you have any more questions, ping me in the nineteenth byte with @ConorO'Brien

Japt, 4 bytes
<âÔx

Try it online!

RProgN, 8 Bytes
~_]k+2/<

Explained
~_]k+2/<
~           # Zero Space Segment
 _          # Convert the input to an integer
  ]         # Duplicate the input on the stack
   k+       # Get the sum of the divisors of the top of the stack
     2/     # Divded by 2
       <    # Is the Input less than the sum of its divisors/2.

Try it online!

R, 32 28 bytes
2*(n=scan())<(x=1:n)%*%!n%%x

Try it online!
Reads n from stdin, returns TRUE for abundant numbers and FALSE otherwise.

C# (Visual C# Interactive Compiler), 48 bytes
x=>Enumerable.Range(1,x).Sum(y=>y<x&x%y<1?y:0)>x

Try it online!
Above is a LINQ based solution that creates a range of numbers slightly larger than the input number. Each element is tested for being a) being a proper divisor of the input and b) being less than the input. If both checks pass, it is added to a sum which is compared to the input.
Below is a a recursive solution that uses an inner function to calculate the summation. While this one seems the most interesting to me, I think the LINQ based approach will not be beat.
C# (Visual C# Interactive Compiler), 60 bytes
x=>{int g(int y)=>(x%y<1?y:0)+(++y<x?g(y):0);return g(1)>x;}

Try it online!
Finally, we have an iterative solution which currently is between the LINQ based approach and the recursive one.
C# (Visual C# Interactive Compiler), 53 bytes
x=>{int y=0,z=0;for(;++y<x;z+=x%y<1?y:0);return z>x;}

Try it online!

PowerShell, 43 bytes
param($n)1..$n|%{$s+=$_*!($n%$_)}
$s-gt2*$n

Try it online!

Common Lisp, 63 bytes
(lambda(n)(>(loop for i from 1 below n if(=(mod n i)0)sum i)n))

Try it online!

F#, 51 bytes
let f n=Seq.filter(fun x->n%x=0){1..n-1}|>Seq.sum>n

Try it online!
Filters out all the numbers that don't divide evenly into n, then sums them and compares them against n.

x86 opcode edited from 179016, 20 bytes
f:      mov ecx, eax       ; copy input number N into ECX
        lea ebx, [eax+eax] ; copy input number 2N into EBX
.a:     cdq                ; clear DX (high word for dividend)
        push eax           ; save original dividend
        div ecx            ; divide DX:AX / CX
        pop eax            ; restore dividend
        test edx, edx      ; if remainder = 0, it divides evenly and is a divisor of N
        jnz .b             ; if not, continue loop
        sub ebx, ecx       ; Subtract divisor from EBX; this simultaneously changes the
                           ; running total, and sets the flags for the return value.
        jc .c              ; If CF=1, the sum of divisors subtracted from N has exceeded
                           ; N, thus N is abundant, and we need to exit the loop now to
                           ; return this result in CF.
.b:     loop .a
.c:                        ; return value: CF=0 (non-abundant), CF=1 (abundant)

Try it online!
Saved from the reverse working

Try it online! (please change to a direct link if you include it) Note that this TIO is not executing exactly the same machine code, as it needs to use 0x66 and 0x67 prefixes to do 16-bit operations, but I still think it's much better than nothing. :-)

@Deadcode Totally changed to 32-bit code and fixed N=1 case with the price that >=0x80000000 fail

32-bit means you can use the hex-dump code I added: Try it online!. Cool that you chose a different way of dealing with input=1. In yours, values of 2^31 or larger can't be handled. In mine, all 32-bit inputs can be handled, making it 3 bytes longer than yours. (And of course, neither of them can handle input=0.)

Whispers v2, 49 bytes
> Input
>> ∤1
>> ∑2
>> 3-1
>> 4>1
>> Output 5

Try it online!
How it works
Whispers is a non-linear language, where numbers don't refer to their numerical value; rather they reference the result of that line in the program. Execution always starts at the last line, and branches through these references.
Program structure tree (green are nilad lines):

Therefore, we start on line 6 by outputting the result of line 5, which is
>> 4>1

which calls lines 4 and 1 and compares the results.
Line 1 is the odd line out in the program: as it starts with a single >, rather than a double >>, it is a nilad line. This means it returns its value interpreted literally. Numbers return their numeric value, and Input is preprocessed at the start of execution, being replaced with the first input on STDIN. Therefore, line 5 checks if the result for line 4 is greater than the input.
Line 4, >> 3-1, subtracts the input from line 3's result, while line 3, >>∑2, sums the values on line 2. Line 2, >> ∤1, generates the list of the input's divisors. Therefore, together, the program generates a list of the input's divisors (including the input), sums them, then subtracts the input (in order to account for the input being included in the list of divisors). Finally, we compare the sum to the input, and output the result.

Zsh, 39 bytes
for ((;++b<$1;t+=$1%b?0:b)):
<<<$[t>$1]

Try it online!

Vyxal, 3 bytes
∆K<

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Lua (71 68 bytes)
Thank to @Deadcode for finding a typo in the byte length and for suggesting shorter code.
function A(n)z=0 for i=1,n-1 do z=z+i*#{[n%i+1]=0}end return z>n end

Welcome to CCGC, and -2 bytes by using z+i*#{[n%i+1]=0} instead of n%i<1 and z+i or z - Try it online!. Also, how did you measure your length? It seems to be 54 bytes even without the function header and footer.

@Deadcode Typo fixed and suggestion included. Thank you for the nice trick with [0]

><>, 47+3 46+3 39+3 = 42 bytes
:l)?!v}l:{:}$%0=*{
 v-$0<
l<+v?=1
)n;>0

Expects the input number to be present on the stack at program start, so +3 bytes for the -v flag. Try it online!
Edit: Golfed 1 byte from 3rd line: -\~0$ => -\:+
Edit 2: Near complete rewrite, previous version:
1v
}/!?)@:{:}+1r~?$r}:}%@:{:
-\:+
?\+l1=
;>0)n

JavaScript ES6, 61 50 bytes
a=>[...Array(a).keys()].reduce((b,c)=>a%c?b:b+c)>a

f=a=>[...Array(a).keys()].reduce((b,c)=>a%c?b:b+c)>a;

console.log(f(11));
console.log(f(12));
console.log(f(17));
console.log(f(18));

45 bytes

Haskell, 34 bytes
a n=n<sum[a|a<-[1..n-1],mod n a<1]

Try it online! Usage example: a 12 returns True.

Pyth, 8 bytes
>s{*MPyP

Try it online here.
>s{*MPyPQQ   Implicit: Q=eval(input())
             Trailing QQ inferred
       PQ    Prime factors of Q
      y      Take the powerset of the above
     P       Discard the last one (i.e. the original set itself)
   *M        Take the product of each list
  {          Deduplicate
 s           Take the sum
>        Q   Is the above greater than Q? Implicit print

APL(NARS) 11 chars, 22 bytes
{⍵<2÷⍨11π⍵}

11π is the function sum divisor, so sum proper divisor would be: {(-⍵)+11π⍵}; test:
  f←{⍵<2÷⍨11π⍵}
  {1=f ⍵:⍵⋄⍬}¨1..100
12  18  20  24  30  36  40  42  48  54  56  60  66  70  72  78  80  84  88  90  96  100

C# (.NET Core), 59, 53 bytes
Without LINQ.
EDIT:  Thanks to ASCII-only for pointing out for(;++k<p;) instead of for(;k<p;k++).
C# (.NET Core), 53 bytes
p=>{int k=0,n=0;for(;++k<p;)n+=p%k<1?k:0;return n>p;}

Try it online!

53

PHP, 63 bytes
function a($x){for($i=1;$i<$x;$i++){$s+=$x%$i?0:$i;}echo$s>$x;}

Usage: a(12);

Forth (gforth), 45 bytes
: f 0 over 1 ?do over i mod 0= i * - loop < ;

Try it online!
Explanation
Loops over every number from 1 to n-1, summing all values that divide n perfectly. Returns true if sum is greater than n
Code Explanation
: f                \ start word definition
  0 over 1         \ create a value to hold the sum and setup up the bounds of the loop
  ?do              \ start a counted loop from 1 to n. (?do skips if start = end)
    over           \ copy n to the top of the stack
    i mod 0=       \ check if i divides n perfectly
    i * -          \ if so, use the fact that -1 = true in forth to add i to the sum
  loop             \ end the counted loop
  <                \ check if n is less than the sum
;                  \ end the word definition

Perl 5 -MList::Util=sum -pa, 32 bytes
$_=$_<sum grep!("@F"%$_),1..$_-1

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Python 3, 45 bytes
lambda n:sum(i for i in range(1,n)if n%i<1)>n

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Factor + project-euler.common, 9 bytes
abundant?

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Builtin, lol

C (clang), 47 bytes
g,o;l(f){for(g=o=f;--o;)g-=f%o?0:o;return g<0;}

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cQuents, 6 bytes
$<U\z$

Try it online!
Outputs 1 for true and 0 for false. If no input given, outputs the 1-indexed sequences of 1/0 for whether each index is an abundant number.
Explanation
        each term in the sequence equals
$       index
 <            <
  U             sum (                           )
   \z                 proper divisors (       )
     $                                  index

Pyt, 7 bytes
ĐĐð⇹\Ʃ<

Try it online!
ĐĐ          implicit input (n); duplicate on stack twice
  ð         get all factors of n
   ⇹\       remove n from list of factors (i.e. reduce list to proper divisors)
     Ʃ      sum proper divisors
      <     is the sum greater than n?; implicit print

Arturo, 26 bytes
$[n]->n<sum chop factors n

Try it

Java 7, 69 68 bytes
Object c(int n){int s=0,i=1;for(;++i<=n/2;s+=n%i<1?i:0);return s>n;}

Explanation:
Object c(int n){                       // Boolean method with integer parameter
   int s = 0,                          // sum initialization
       i = 1;                          // counter initialization (starts at 2: i=1 and ++i becomes i=2)
   for(; ++i <= n/2; s += n % i < 1    // loop while (i <= n/2)
                           ? i         //   if (n % i == 0)
                           : 0);       //     r = r + i
   return s > n;                       // return sum > input
 }

Test code:
Try it here.
class M{
  static Object c(int n){int s=0,i=1;for(;++i<=n/2;s+=n%i<1?i:0);return s>n;}

  public static void main(String[] a){
    for(int i = 0; i <= 100; i++){
      System.out.print(i+": "+c(i) + ";  ");
    }
  }
}

Output:
0: false;  1: false;  2: false;  3: false;  4: false;  5: false;  6: false;  7: false;  8: false;  9: false;  10: false;
11: false;  12: true;  13: false;  14: false;  15: false;  16: false;  17: false;  18: true;  19: false;
20: true;  21: false;  22: false;  23: false;  24: true;  25: false;  26: false;  27: false;  28: false;  29: false;
30: true;  31: false;  32: false;  33: false;  34: false;  35: false;  36: true;  37: false;  38: false;  39: false;
40: true;  41: false;  42: true;  43: false;  44: false;  45: false;  46: false;  47: false;  48: true;  49: false;
50: false;  51: false;  52: false;  53: false;  54: true;  55: false;  56: true;  57: false;  58: false;  59: false;
60: true;  61: false;  62: false;  63: false;  64: false;  65: false;  66: true;  67: false;  68: false;  69: false;
70: true;  71: false;  72: true;  73: false;  74: false;  75: false;  76: false;  77: false;  78: true;  79: false;
80: true;  81: false;  82: false;  83: false;  84: true;  85: false;  86: false;  87: false;  88: true;  89: false;
90: true;  91: false;  92: false;  93: false;  94: false;  95: false;  96: true;  97: false;  98: false;  99: false;  100: true;

Sorry, I was looking at the main code in that part but the golfed code above. Reading failure.

I am most likely overlooking something obvious, though I am curious: Why did you choose to use Object as opposed to bool?

@JonathanFrech In Java it's boolean, so it's 1 byte longer than Object :) Although nowadays I would just use a Java 8+ lambda, so Object c(int n) would be n->.

Scala, 39 bytes
(n:Int)=>(1 until n filter{n%_<1}sum)>n

Javascript, 47 Bytes

a=>{s=d=0;while(a>d){s+=a%d?0:d;d++}return s>d}

f=a=>{s=d=0;while(a>d){s+=a%d?0:d;d++}return s>d};
console.log(f(11));
console.log(f(12));
console.log(f(15));
console.log(f(18));

Quite straight forward: check all numbers smaller than the input, if the input is divisible by them add them to the sum, and return whether the sum is bigger than the input

C, 68 bytes
int s,x;main(i){scanf("%d",&x);while(++i<x)s+=x%i^0?0:i;return s>x;}

Welcome to the site! I think you could save some bytes by taking input from function arguments, rather than implementing main with scanf. For example int f(int s,int x){while(++i<x)s+=x%i^0?0:i;return s>x;} Also, I'm not an expert at C golfing, but I think some compilers default to int functions with int arguments, so you could do f(s,x){while(++i<x)s+=x%i^0?0:i;return s>x;}, but you'd have to find the right compiler and test it first.

Racket, 46 bytes
That's 46 bytes including the (require math).
(λ(n)(>(-(sum(divisors n))n)n))

The function produces booleans.
Try it online!
It's pretty simple, uses the divisors function to produce a list of all divisors of n. Since the list includes n itself, we need to remove it after summing.

Hello and welcome to PPCG. Nice first post; however, I would think that you ought to include the byte count for your (require math) import.

Sorry! I was unsure about what exactly to add to the byte count.

No problem. Usually, one counts everything the function or program truly needs -- excluding boiler code only written for testing. However, on edge cases one sometimes has to resort to the current consensus -- written and talked about on meta. In practice, you will often be notified by comment if someone disagrees with your byte count.

Java 8, 66 bytes
n->java.util.stream.IntStream.range(1,n).filter(i->n%i==0).sum()>n

Try it online!

Explanation
n->                                       // define a lambda function
    java.util.stream.IntStream.range(1,n) // create a stream of integers from 1 to n
    .filter(i->n%i==0)                    // filter out non-factors
    .sum()                                // sum the factors
>n                                        // compare the sum to the original number n
                                          // implicit return

Edit: Looks like someone already did this 2 years ago and got 1 byte shorter

Javascript,72bytes
var n=prompt(),div=0,i=1;while(++i<=n/2){div+=n%i?0:i}console.log(div>n)

As @SriotchilismO'Zaic points out, this appears to be a snippet which is not compliant with our current I/O consensus. Please fix your post accordingly.

As @JonathanFrech specified, modified the changes and further refined the code.

TI-BASIC (TI-84), 23 bytes
:2Ans<sum(seq(Ans/Xnot(remainder(Ans,X)),X,1,Ans,1

This answer is very similar to my answer here.
Input is in Ans.
Output is in Ans and is automatically printed out when the program completes.
Explanation:
(TI-BASIC doesn't have comments, so just assume that ; makes a comment)
:2Ans=sum(seq(Ans/Xnot(remainder(Ans,X)),X,1,Ans    ;Full program

 2Ans                                               ;double the input
          seq(                                      ;generate a list
                                         X,          ;using the variable X,
                                           1,        ;starting at 1,
                                             Ans     ;and ending at the input
                                                     ;with an implied increment of 1
              Ans/X                                 ;from the input divided by X
                   not(                ),           ;multiplied by the negated result of
                       remainder(Ans,X)              ;the input modulo X
                                                     ;(result: 0 or 1)
      sum(                                          ;sum up the elements in the list
     <                                              ;is the sum greater than double the input?

Example:
12
           12
prgmCDGF4
            1
6
            6
prgmCDGF4
            0

Note:  The byte count of a program is evaluated using the value in [MEM]>[2]>[7] (36 bytes) then subtracting the length of the program's name, CDGF2, (5 bytes) and an extra 8 bytes used for storing the program:
36 - 5 - 8 = 23 bytes

I counted 50 characters, so presumably 50 bytes... what am I missing?

@Adam all of the functions in TI-BASIC are 1-byte tokens.  Therefore, sum(, seq(, remainder(, and not(  would all be considered as 1 byte each.  Ans is also a 1-byte token.

Ah, okay. I see now

Actually, I should clarify that the remainder( function is a 2-byte token, not 1-byte.

Kotlin, 37 bytes
{i->(2..i-1).filter{i%it==0}.sum()>i}

Expanded:
{ i -> (2..i - 1).filter { i % it == 0 }.sum() > i }

Explanation:
  {i->                      start lambda with parameter i
    (2..i-1)                create a range from 2 until i - 1 (potential divisors)
      .filter{i%it==0}      it's a divisor if remainder = 0; filter divisors
        .sum()              add all divisors
          >i                compare to the original number

Try it online!

TI-BASIC, 21 bytes
seq(I,I,1,A-1
A<sum(Ansnot(fpart(A/Ans

TI-BASIC usually has a builtin for this kind of thing, but nothing fits that well here.  Takes input from the A variable and outputs 1 or 0 if the number is or isn't abundant.
seq( generates a list of numbers from a function.  Here it executes the function I on the I variable from one to one less than the input.  Then working backward from the next line, we divide our input by the list we generated (stored in Ans) and take the fractional part so we have a new list of remainders for each potential divisor.  We then build a list of booleans by taking not( on the remainder list and multiply that by the original list in Ans implicitly to get a list of proper divisors.  Then we just sum the list of proper divisors and compare that to the input.
This was 24 bytes, golfed down to 21 using not( instead of *(0=

tinylisp, 78 bytes
(load library
(q((N)(l N(sum(filter(partial divides?(q D)(list(q _)N))(range N

Try it online!
Start with a list of each D from 0 through N - 1; filter it for the values of D that exactly divide N. Sum the results. Iff N is less than that sum, N is abundant.

Excel, 46 bytes
=LET(s,SEQUENCE(A3-1),SUM((MOD(A3,s)=0)*s)>A3)

Works for numbers up to 2^20 + 1 (1,048,577).
This 90 byte version works for numbers up to (2^20 + 1) ^ 2 - 1 (1,099,513,724,928).
=LET(s,SEQUENCE(SQRT(A2)),m,(MOD(A2,s)=0)*s,g,IFERROR(A2/m,0),SUM(m,g)-A2-SUM((m=g)*m)>A2)

Thunno 2, 4 bytes
FṫS<

Attempt This Online!
Explanation
FṫS<  # Implicit input      ->  12
F     # Factors             ->  [1, 2, 3, 4, 6, 12]
 ṫ    # Tail remove         ->  [1, 2, 3, 4, 6]
  S   # Sum the list        ->  16
   <  # Compare with input  ->  1
      # Implicit output

authors: J. А. del Rio - слава СССР

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/