user13928: What is the shortest game with the largest material imbalance? This of course means that one side will have 9 queens, 2 rooks, 2 bishops, 2 knights, and a king, and a bare king on the other side. From some experimenting, I got this:
[FEN ""]

1.h4 g5 2.hxg5 h6 3.gxh6 Rh7 4.g4 Rg7 5.h7 Rg6 6.h8=Q Bh6 7.g5 f5 8.f4 e6 9.e4 d5 10.d4 c5 11.c4 b5 12.b4 a5 13.a4 Qe7 14.exf5 Bb7 15.fxe6 Bc6 16.cxd5 Bb7 17.bxc5 Bc8 18.axb5 Bb7 19.c6 Bc8 20.Rxa5 Ra7 21.b6 Rd7 22.Qe5 Nf6 23.b7 Kf8 24.c7 Rd6 25.f5 Kg8 26.Nc3 Bd7 27.c8=Q+ Kh7 28.Qc4 Nc6 29.b8=Q Be8 30.Bd2 Qf8 31.e7 Qg7 32.Bg2 Bf7 33.e8=Q Re6 34.dxe6 Bg8 35.d5 Nb4 36.d6 Nc6 37.d7 Nb4 38.d8=Q Nc6 39.e7 Nb4 40.Qdd7 Na6 41.Qed8 Nc5 42.e8=Q Nd3+ 43.Qcxd3 Ng4 44.f6 Ne3 45.f7 Nf5 46.f8=Q Qf7 47.Qfb4 Rg7 48.g6+ Kh8 49.gxf7 Ne7 50.Q3d6 Kh7 51.Q6xe7 Kh8 52.f8=Q Kh7 53.Qed5 Rf7 54.Bh3 Kh8 55.Q7xf7 Bg7 56.Ke2 Kh7 57.Ke3 Kh6 58.Kf4 Kh7 59.Q8xg8+ Kh6 60.Qgxg7#  0-1

This can of course be improved though, as I was essentially moving pieces around randomly.

user15806: I got this, which is much shorter.
[FEN ""]

1. a4 a5 2. b4 b5 3. c4 c5 4. d4 d5 5. e4 e5 6. f4 f5 7. g4 g5 8. h4 h5 9. bxa5 Bd6 10. axb5 Be6 11. dxc5 Nf6 12. cxd5 O-O 13. fxe5 Rf7 14. exf5 Ra6 15. gxh5 Rb7 16. hxg5 Rbb6 17. bxa6 Nc6 18. axb6 Qe7 19. dxc6 Kh8 20. cxd6 Ng8 21. a7 Qf6 22. gxf6 Ne7 23. b7 Bd7 24. e6 Ng8 25. c7 Kh7 26. f7 Kg7 27. e7 Kf6 28. h6 Ke5 29. f6 Kf5 30. a8=Q Kg6 31. b8=Q Ba4 32. c8=Q Bb5 33. f8=Q Bc4 34. d7 Bd5 35. d8=Q Be4 36. f7 Bf5 37. e8=Q Bd7 38. h7 Bc6 39. h8=Q Bb7 40. Qcxb7 Kf5 41. fxg8=Q#

It's one move above the lower bound, but I feel that this is optimal.

As @Evargalo suggested, here is @user1583209's proof of optimality from their comments.
"I have a proof now: Start by labeling the columns/files by numbers, i.e. a=1, b=2...h=8. Then for each pawn calculate the sum of the respective files. So in the initial position this sum is 1+2+3...+8=36. Now, if a pawn captures, it changes its file by "1" (either plus or minus 1). In order to achieve a solution in 40 moves, we have to queen the pawns on each of the squares on the 8th rank. This means that in the final position the sum of files (for pawns+promoted pawns now) needs to be 36 again. However we also need to do 15 captures with these pawns.
Each capture changing the file sum by 1, it is impossible to end up with 36 as sum, basically because 15 is an odd number.  (If you have x captures to the left and y captures to the right you end up with equations: x-y=0 and x+y=15 which has no integer solution) That proves that it is impossible to achieve a solution in 40 moves and the solution with 41 moves given by orlp is optimal."

13928: <p>What is the shortest game with the largest material imbalance? This of course means that one side will have 9 queens, 2 rooks, 2 bishops, 2 knights, and a king, and a bare king on the other side. From some experimenting, I got this:</p>

<pre><code>[FEN ""]

1.h4 g5 2.hxg5 h6 3.gxh6 Rh7 4.g4 Rg7 5.h7 Rg6 6.h8=Q Bh6 7.g5 f5 8.f4 e6 9.e4 d5 10.d4 c5 11.c4 b5 12.b4 a5 13.a4 Qe7 14.exf5 Bb7 15.fxe6 Bc6 16.cxd5 Bb7 17.bxc5 Bc8 18.axb5 Bb7 19.c6 Bc8 20.Rxa5 Ra7 21.b6 Rd7 22.Qe5 Nf6 23.b7 Kf8 24.c7 Rd6 25.f5 Kg8 26.Nc3 Bd7 27.c8=Q+ Kh7 28.Qc4 Nc6 29.b8=Q Be8 30.Bd2 Qf8 31.e7 Qg7 32.Bg2 Bf7 33.e8=Q Re6 34.dxe6 Bg8 35.d5 Nb4 36.d6 Nc6 37.d7 Nb4 38.d8=Q Nc6 39.e7 Nb4 40.Qdd7 Na6 41.Qed8 Nc5 42.e8=Q Nd3+ 43.Qcxd3 Ng4 44.f6 Ne3 45.f7 Nf5 46.f8=Q Qf7 47.Qfb4 Rg7 48.g6+ Kh8 49.gxf7 Ne7 50.Q3d6 Kh7 51.Q6xe7 Kh8 52.f8=Q Kh7 53.Qed5 Rf7 54.Bh3 Kh8 55.Q7xf7 Bg7 56.Ke2 Kh7 57.Ke3 Kh6 58.Kf4 Kh7 59.Q8xg8+ Kh6 60.Qgxg7#  0-1
</code></pre>

<p>This can of course be improved though, as I was essentially moving pieces around randomly.</p>
 15806: <p>I got this, which is much shorter.</p>
<pre><code>[FEN &quot;&quot;]

1. a4 a5 2. b4 b5 3. c4 c5 4. d4 d5 5. e4 e5 6. f4 f5 7. g4 g5 8. h4 h5 9. bxa5 Bd6 10. axb5 Be6 11. dxc5 Nf6 12. cxd5 O-O 13. fxe5 Rf7 14. exf5 Ra6 15. gxh5 Rb7 16. hxg5 Rbb6 17. bxa6 Nc6 18. axb6 Qe7 19. dxc6 Kh8 20. cxd6 Ng8 21. a7 Qf6 22. gxf6 Ne7 23. b7 Bd7 24. e6 Ng8 25. c7 Kh7 26. f7 Kg7 27. e7 Kf6 28. h6 Ke5 29. f6 Kf5 30. a8=Q Kg6 31. b8=Q Ba4 32. c8=Q Bb5 33. f8=Q Bc4 34. d7 Bd5 35. d8=Q Be4 36. f7 Bf5 37. e8=Q Bd7 38. h7 Bc6 39. h8=Q Bb7 40. Qcxb7 Kf5 41. fxg8=Q#
</code></pre>
<p>It's one move above the lower bound, but I feel that this is optimal.</p>
<hr />
<p>As @Evargalo suggested, here is @user1583209's proof of optimality from their comments.</p>
<p>&quot;I have a proof now: Start by labeling the columns/files by numbers, i.e. a=1, b=2...h=8. Then for each pawn calculate the sum of the respective files. So in the initial position this sum is 1+2+3...+8=36. Now, if a pawn captures, it changes its file by &quot;1&quot; (either plus or minus 1). In order to achieve a solution in 40 moves, we have to queen the pawns on each of the squares on the 8th rank. This means that in the final position the sum of files (for pawns+promoted pawns now) needs to be 36 again. However we also need to do 15 captures with these pawns.
Each capture changing the file sum by 1, it is impossible to end up with 36 as sum, basically because 15 is an odd number.  (If you have x captures to the left and y captures to the right you end up with equations: x-y=0 and x+y=15 which has no integer solution) That proves that it is impossible to achieve a solution in 40 moves and the solution with 41 moves given by orlp is optimal.&quot;</p>
