A New Year's Math Riddle Here's a fun and clean riddle for the New Year! I found the material for it on Krabbe's site. I will provide the link when it is solved.
How many times can this position be repeated, with both sides helping, before either side can claim a draw repetition? No cheating, please!

Addendum 1/7/20201: With the acceptance of the answer, here is the link to Krabbe's AD magazine article &quot;TWEEëNTWINTIG FOTO'S,&quot; which translates to &quot;Twenty Two Photos.&quot;
The board layout can be repeated at most 11 times with White having the move, the 12th will be a draw.
[FEN &quot;rnbqkbnr/1pp2pp1/4p3/p2pP2p/P6P/8/1PPP1PP1/RNBQKBNR w KQkq d6 0 1&quot;]

1. Be2 {White can no longer capture en passant.} Be7 2. Bf1 Bf8 {2nd time} 3. Be2 Be7 4. Bf1 Bf8 {3rd time}
5. Rh2 {White can no longer 0-0} Be7 6. Rh1 Bf8 {4th time} 7. Rh2 Be7 8. Rh1 Bf8 {5th time}
9. Ra2 {White can no longer 0-0-0} Be7 10. Ra1 Bf8 {6th time} 11. Ra2 Be7 12. Ra1 Bf8 {7th time}
13. Be2 Rh7 {Black can no longer 0-0} 14. Bf1 Rh8 {8th time} 15. Be2 Rh7 16. Bf1 Rh8 {9th time}
17. Be2 Ra7 {Black can no longer 0-0-0} 18. Bf1 Ra8 {10th time} 19. Be2 Ra7 20. Bf1 Ra8 {11th time}

However (emphasis mine):

9.2.2  Positions are considered the same if and only if the same player has the move, pieces of the same kind and colour occupy the same squares and the possible moves of all the pieces of both players are the same.

So:
[FEN &quot;rnbqkbnr/1pp2pp1/4p3/p2pP2p/P6P/8/1PPP1PP1/RNBQKBNR w KQkq d6 0 1&quot;]

1. Bd3 Bc5 2. Be2 Bf8 3. Bf1

is not a repetition of the original position (it's Black's turn now), although it is the same board layout.
Therefore: the position can be repeated 10 times (excluding the initial position), multiply this by 2 (2 more times the layout can be repeated with Black's turn) and add 1 (the initial position) + the final repetition on which the draw is claimed = 22 times.