I utilized idea Qb8, so here is full 94 solution:
1.b3 g6 2.b4 g5 3.b5 g4 4.b6 g3 5.hxg3 axb6 6.a3 h6 7.a4 h5 8.a5 h4 9.a6 h3 10.a7 h2 11.Rxh2 Rxa7 12.Rh3 Ra6 13.Rh4 Ra5 14.Rh5 Ra4 15.Rh6 Ra3 16.Rh7 Ra2 17.Rxa2 Rxh7 18.Ra3 Rh6 19.Ra4 Rh5 20.Ra5 Rh4 21.Ra6 Rh3 22.Ra7 Rh2 23.Ra8 Rh1 24.c3 f6 25.c4 f5 26.c5 f4 27.c6 f3 28.exf3 dxc6 29.d3 e6 30.Nd2 Ne7 31.Nb3 Ng6 32.Nd4 Ne5 33.Nf5 Nc4 34.Nh6 Na3 35.Bd2 Be7 36.Be3 Bd6 37.Bf4 Bc5 38.Bg5 Bb4+ 39.Ke2 Kd7 40.Bf6 Bc3 41.Be7 Bd2 42.Nf7 Nc2 43.Nh8 Na1 44.g4 b5 45.g5 b4 46.g6 b3 47.g7 b2 48.g8=N b1=N 49.g3 b6 50.g4 b5 51.g5 b4 52.g6 b3 53.g7 b2 54.f4 c5 55.f5 c4 56.f6 c3 57.f7 c2 58.f8=N+ Kc6 59.Kf3 c1=N 60.Qe2 Qd7 61.Kg4 Kb5 62.f3 c6 63.f4 c5 64.f5 c4 65.f6 c3 66.f7 c2 67.Qf3 Qc6 68.Bg2 Bb7 69.d4 e5 70.d5 e4 71.d6 e3 72.d7 e2 73.d8=N e1=N 74.Ne2 Nd7 75.Ng3 Nb6 76.Ne4 Nd5 77.Ng5 Nb4 78.Nge6 Nbd3 79.Nc7+ Kc4 80.Kf5 Nf2 81.Ne8 Nd1 82.Qf4+ Kd3 83.Bf3 Qc5+ 84.Ke6 Bc6 85.Be4+ Ke2 86.Qe5 Qd4 87.Qd6 Bd5+ 88.Kd7 Qe3 89.Kc8 Kf1 90.Qc7 Qf2 91.Qb8 Qg1 92.Bf5 Bc4 93.Bg6 Bb3 94.Bh7 Ba2 *
> I think, the point is impossibility to keep ALL 32 pawns and pieces of both sides to make needed 6 * 32 = 192 half-moves.
There is no need to keep all pawns and pieces --- the only aim is maximum number of moves. However, as you see, loosing pawns or pieces leads to lower number of moves. Subtle, non-evident point is that sometimes loosing pawns or pieces do NOT leads to lower number of moves.
94 moves - pretty impressive. The lost half-moves are b7, c7 and f2, g2 pawns which made only 5 half-moves each instead of 6. Is there a way to improve a result?
To #31. Instead of 22. Ra7 try 22. g2xh3 and from other side as well and pass that pawns.
Sorry, previous was to #29
Another 94 moves game with all pieces on 8-th row and no pieces on 7-th row:
http://ru.lichess.org/mItYVGQK96 moves still seems to be impossible.
The problem seems to be from the beginning: b2 and g7 pawns (as well as d2 and e7 pawns in my example later) are captured on 6-th row, making only 4 half-moves instead of 5. IF they DID make 5 moves, the puzzle could be solved, but I think it is impossible due to the pawn structure.
So this lacking 4 half-moves (from 188 to 192) are exactly the result, that 4 pawns: b2, g7, d2 and e7 make 1 half-move each less then needed. I think, it can not be improved.
This does not SEEM to be impossible.
It IS impossible.
Since then I have shown this problem to the scribes, and here is a PROOF that 94 is indeed the maximum ( I don't give a link because I'm not advertising here other chess sites ):
Maximum without collisions: 104 moves (pawns 8*6 + pieces 8*7)
Now we lose moves because of collisions - not all the units get to make maximum moves. Captures are also necessary, meaning some units don't get to make all their moves - i.e. we lose halfmoves. It remains to show each side loses halfmoves in a way that results in the fewest total moves lost.
The 4 rooks colliding with each other must lose 2 halfmoves, e.g. wRa1-a2-...-a7, bRa8xa7-...-a1, same for h-file.
The pawns, in order to get past each other, must make captures. Unfortunately there are lots of ways to do this. Each capture at minimum loses 2 halfmoves (since any piece captured on its 6th rank loses the 2 remaining moves.) The apparently optimal way to do this is for instance wPa2-...-a6, bPb7xa6 and all of the a- and b-pawns are clear; repeat for each pair of files, resulting in 8 lost halfmoves from 4 pawn captures.
Note that this capture scheme also means that the edge rooks need to capture 2 edge pawns still, in order to reach the 8th rank. So this is another 2 lost halfmoves that the pawns could have made.
Our theoretical maximum also relies on all units reaching their last ranks. Obviously impossible since there are 16 units per side and only 8 squares on the last rank. 32 units - 4 captured pawns - 2 captured rooks - 2 captured edge pawns = 24 units and 16 squares to fill. Therefore at least 8 more halfmoves must be wasted.
Our total wastage is therefore 2+8+2+8 = 20 halfmoves, which is at best 10 full moves. So 104 - 10 = 94 moves is the theoretical upper bound.
To #38: "24 units and 16 squares to fill" - in my example there are only 16 units to fill all 16 squares, and it is still 94 moves, so reducing the amount of units does not necessarily reduce the amount of moves.
I think simpler explanation, that according to conditions of puzzle each unit should make 6 half-moves in average (192/32 = 6). It is impossible that EACH unit make these 6 half-moves as it would be too "crowdy". Other way is when some units (some pawns) make 5 moves, while other units (some pieces) make 7 moves. Note that each 5-moves unit refer only to one 7-moves unit (each captured pawn frees only 1 place in 8-th row for a piece). So, when the pawn HAS to be captured on 6-th row making ONLY 4 half-moves, it can not be compensated by one 7-moves piece (7+4 = 11, not 12). And this capture of the pawn on 6-th row can not be avoided.
I'm the one who posted that solution on the other chess site, so I'll address that issue.
I took a look at your example of 16 units to fill 16 squares - and it's really the same 24 units again. Or rather, it would be if they weren't captured after reaching their final squares. Either way (having been captured or just left on the 7th rank), they lose one halfmove each - nothing changes.
Basically that section of the proof is explaining not all the uncaptured pieces can make their maximum moves; only 16 at most can, and therefore we lose 24-16 = 8 halfmoves, regardless of whether they are still on the board at the end or not.
Your proof shows clearly that 192/2 = 96 is impossible. What mine shows is rather that 94 is the greatest lower bound for game length (and justified by a proof game, e.g. yours.) (This was also the question asked on the other site, whether 94 was really the maximum and not say 95.)