【Analysis】When $M=N$, the first player A must lose; when $M$ is not equal to $N$, A can use a unique strategy to return to the starting point of the board, making B go first, and B will lose. There are 10 cases where A will lose, and $100-10=90$ cases where A will win.

【Solution】Assume A goes first, and B goes second,
First, it is easy to see that when $M=N=1$, A must lose. Below, we will prove by induction that when $M=N$, A must lose. The case $M=N=1$ is obvious.

Assume that when $M=N \leqslant K$, A must lose, then when $M=N=K+1$, if A takes the first step in any direction for $X$ steps, B will take $X$ steps in the other direction.

If $X=K+1$, it is easy to see that the right upper corner has been reached, so A has already lost. Otherwise, if $X < K+1$, A can take the first step directly along the direction of length $M$ for $M-N$ steps, thus transforming the game into a new $(N, N)$ board, and it is now B's turn to move. By the above conclusion, B must lose, meaning A must win.

In summary, for A to win, it is only necessary that $M$ is not equal to $N$.
Thus, the number of such pairs of positive integers is $10 \times 10-10=90$.
Therefore, the answer is 90.
10. (10 points) A rectangular chessboard composed of unit squares of size $m \times n$ (where $m, n$ are positive integers not exceeding 10), has a chess piece placed in the unit square at the bottom-left corner. Two players, A and B, take turns moving the piece. The rules are: move up any number of squares, or move right any number of squares, but you cannot move out of the board or not move at all. The player who cannot make a move loses (i.e., the first player to move the piece to the top-right corner wins). How many pairs of positive integers $(m, n)$ allow the first player A to have a winning strategy? $\qquad$