Antipov $M$.

In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of the figure $\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). We will call such a placement unsuccessful if the sum of the numbers in the four cells of the figure is not equal to 0. Find the smallest possible number of unsuccessful placements.
Evaluation. Let's show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Let the numbers in the outer cells of the cross be \(a, b, c, d\), and in the central cell be \(e\); denote by \(S\) the sum of all these five numbers. Suppose all placements in the cross are successful. Then \(S-a=S-b=S-c=S-d=0\), from which \(a=b=c=d\). Therefore, \(e + 3a = 0\), that is, \(e = -3a = \pm 3\), which is impossible. Thus, in each of the 36 "crosses" (with centers in all non-edge cells), there is an unsuccessful placement. It is clear that each placement is contained in no more than one cross; therefore, there are at least 36 such placements.

![](https://cdn.mathpix.com/cropped/2024_05_06_388b3f8bbd0a17840215g-11.jpg?height=569&width=554&top_left_y=2357&top_left_x=755)

An example of a placement where the number of unsuccessful placements is 36 is shown in the figure (the sign of the corresponding number is indicated in each cell). It is easy to verify that in each cross, there is exactly one unsuccessful placement, and all placements adjacent to the long side of the board's edge are successful.

## Answer

36.