, 2011

First, notice that player two can always ensure that $A \le 6$. The squares of the grid can be partially tiled with $2 \times 1$ tiles, as shown below. Whenever a $1$ is placed inside a tile, player two can ensure that it also contains a $0$. As every $3 \times 3$ square contains three complete tiles, it will have at least three zeros in it so, $A \le 6$.

Label each individual cell with the number of $3 \times 3$ squares that contain it. By going for the highest-placed numbers, player $1$ can get a total of $45$, which spread across $9$ squares gives exactly $5$. Let player $1$ start by placing a $1$ in the middle, and wlog, player two plays in the left side of the board. By placing a $1$ directly right of the middle (value $6$), player two is forced to take the right-most cell in the middle row (else player $1$ will, and there will exist a $3 \times 3$ square with three $1$s and $6$ empty blocks). By taking the block directly below the middle (value $6$), player one leaves the other stranded: if he does not place a $0$ in the bottom-rightmost $3 \times 3$ square, player $1$ will get six $1$s in it. If he does, player $1$ will take the last square of value $6$, and will finish with cell totals greater than $45$, which will give $A \ge 6$. Hence $A = 6$.