Consider an $8 \times 9$ chessboard. How many $1 \times 2$ dominoes (rotations allowed) can be placed if the dominoes $(k, 9-k)-(k, 10-k)$ for $2 \leqslant k \leqslant 7$ have already been placed?
34. Consider the set $A$ consisting of the cells $(1,9),(8,1)$ and the cells located in the top left of the set formed by these two cells and the already placed dominoes. Similarly, we denote $B$ as the analogous set in the bottom right. Any domino other than the six initial ones is entirely contained within one of these two sets.

We color the cells like a chessboard. In $A$, there are 17 black cells and 14 white cells. Since a domino covers an equal number of black and white cells, there can be at most 14 dominos entirely contained within $A$. The same applies to $B$. Therefore, there are at most $6+14+14=34$ dominos in total.