8.117 There are 9 boxes arranged in a $3 \times 3$ square and an ample supply of black and white stones. Two players, A and B, play a game of placing stones: each time, one player places 3 stones in the same row or the same column of 3 boxes, and these 3 stones do not have to be the same color. No box can contain stones of different colors. If a player places 1 white stone in a box that already contains a black stone, then both the white stone and 1 black stone are removed from the box. The game ends when the central and the four corner boxes each contain 1 black stone, and the other boxes are empty. If at some stage of the game, $x$ boxes each contain 1 black stone and the rest of the boxes are empty, find all possible values of $x$.
[Solution] Clearly, $x$ can be $3,6,9,4$ (place a row of black stones first, then place a column of stones, among which 2 are black and 1 is white, making the white stone cancel out 1 of the previously placed black stones), 5 (place two rows of black stones first, then place a column of stones, among which 2 are white and 1 is black, making the two white stones cancel out 2 of the previously placed black stones).

Below, we prove that $x$ cannot take the values $1,2,7,8$. Note that, whether $x$ is 1 or 2, it is always possible to select two rows and two columns from the $3 \times 3$ square such that the 4 boxes at the intersections of these two rows and two columns contain only 1 black stone, with the other 3 boxes being empty. Similarly, when $x$ is 7 or 8, it is also possible to select 4 such boxes, where 3 boxes each contain 1 black stone and the other 1 box is empty.

Correspond each black stone to 1 and each white stone to -1, and only consider the 4 selected boxes. It is easy to see that after each placement of stones, the sum $S$ of the numbers corresponding to these 4 boxes increases by 2, 0, or -2, i.e., the parity of $S$ remains unchanged. Since the boxes are initially all empty, $S$ is always even. However, in the 4 selected boxes, there are 1 or 3 black stones, corresponding to an odd sum $S$, which is a contradiction. This shows that $x$ cannot take the values $1,2,7,8$.