Note: My frequent edits are to either upgrade what I have found or to fix an illegal position or a flaw.
It is a known record for most number of mutual checks in a row. This made me wonder what the most number of possible double checks in a row is. Only one side can deliver a double check of course.
I have four catergories that I have come up with. Promoted pieces allowed and not allowed. That means that the position can’t start out with promtoed piece’s, but it is allowed once the position starts. The first move/double check is permitted to be a promotion, however.
For each of these are two catergories-forced and unforced. Unforced is where the checked player willing moves into the next square where the double check will occur. Forced is where they have no other options.
If checkmate occurs, the mating move only counts if it was a double check.
Three-fold repetition and the 50-move apply, i.e a position can only come two times. It is a draw if three occurs. (I would do this on the last move if ainwere you. It would stil technically count.
No infinite loops allowed, even they are possible.
The position must be legally reachable and contain only legap move.
If a game is forced, and it beats the record for unforced, it is allowed to contend for both titles.
I’ll give you my reasearch on what I could find. The question here is this: Can you find more?
No Promotion, Unforced-6 Double Checks
[FEN "4K3/8/8/3p4/k1p1N3/1p4N1/P5BB/RRQ5 w - - 0 1"] 1. axb3+ Kb5 2. bxc4+ Kc6 3. cxd5+ Kxd5 4. Nc3+ Kd6 5. Nf5+ Kc5 6. Na4#
Update: I decided to take a look at the category that I ommited. I have now decided to re-enter it after being surprisedn at what I found.
No Promotion, Forced-5 Double Checks
[FEN "kb1r4/P1p5/1P2P3/6NB/7B/8/8/RRQ4K w - - 0 1"] 1. axb8=Q+ Kxb8 2. bxc7+ Kc8 3. cxd8=Q+ Kxd8 4. Nf7+ Ke8 5. Nd6+ Kf8 *
Promotion, Unforced-9 Double Checks
[FEN "K5b1/5p2/4p3/3pR3/k1p1N1N1/1pQ5/P1B5/RR1R1RRR w - - 0 1"] 1. axb3+ Kb5 2. bxc4+ Kc6 3. cxd5+ Kd7 4. dxe6+ Ke8 5. exf7+ Kf8 6. fxg8=Q+ Kxg8 7. Nh6+ Kh7 8. Nf6+ Kh8 9. Nf7# *
Promotion, Forced-8 Double Checks
[FEN "K4Bb1/5p2/4p3/3p4/k1p3N1/1pR5/P1B5/RQ1RRRRR w - - 0 1"] 1. axb3+ Kb5 2. bxc4+ Kc6 3. cxd5+ Kd7 4. dxe6+ Ke8 5. exf7+ Kxf8 6. fxg8=Q+ Kxg8 7. Nh6+ Kh8 8. Nf7#
This is now outdated. but I’m keeping it here for asthetic purposes. It’s a beautiful, intricate, and well oiled-machine!
[FEN "3B1Q2/2N2N1R/3N1B2/N3N1B1/5N1B/3B2N1/2K2k2/R7 w - - 0 1"] 1. Nh1+ Ke3 2. Ng2+ Kd4 3. Nf3+ Kc5 4. Ne4+ Kb6 5. Nd5+ Ka7 6. Nc6+ Kb7 7. Nfd6+ Kxc6
Good luck finding more!
SMALL UPDATE: As Hauke Reddmann noted in a comment, Ba1 Bb1 Rb2 - Kc3 is a well-known mechanism. However, he does not state where he know it from. But luckily, I have managed to track down what is probally the orginal problem in which it appears. It comes from The Problemist of course. I found in in the yacpdb chess problem database.
[Title "Стојнић, Драган Бабић, Миломир, The Problemist 2004-05, #13"] [FEN "3q1nKB/R1P1kPRB/N3p3/1p1n2p1/2r2p2/1p2b3/P2pb2N/3r4 w - - 0 1"] 1. c8=N+ Kf6 2. Rg6+ Kf5 3. Rf6+ Ke5 4. Rf5+ Ke45. Re5+ Kd4 6. Re4+ Kd3 7. Rd4+ Kc3 8. Rd3+ Kc2 9. Rc3+ Kb2 10. Rc2+ Kb1 11. Rb2+ Ka1 12. Rb1+ Kxa2 13. Nb4#
However it is not entirely forced because of the possibility of 10. Ka3 by black. I see no way to cover that sqaure without interferring with the sequence, at least so far. Otherwise we could have 13-more than Evargalo's sweet 12.
Another One: As Evargalo said in a comment, this mechanic is probably from a century ago. The problem above is from 2004. And indeed it is from a a century ago-quite literally-from what I found in this forum on Red Hot Pawn.
[Title "Study by A. White, 1919, #12"] [FEN "2q5/2pp4/3pr3/4pb2/K1p2pn1/2bn1kp1/3pr1R1/6BB w - - 0 1"] 1. Rf2+ Ke3 2. Rf3+ Ke4 3. Re3+ Kd4 4. Re4+ Kd55. Rd4+ Kc5 6. Rd5+ Kc6 7. Rc5+ Kb6 8. Rc6+ Kb7 9. Rb6+ Ka7 10. Rb7+ Ka8 11. Ra7+ Kb8 12. Ra8#