3. Suppose $k$ squares have been cut. The intersected diagonals form a graph $G$ whose vertices correspond to the vertices and centers of the $k$ cut squares. This graph has $4k$ edges. To prevent the board from falling apart, the graph must not contain any cycles, so the number of vertices must be at least $4k+1$.

Since none of the edge squares can be cut, all vertices of the graph $G$ belong to the inner $6 \times 6$ square. Moreover, since the graph has an even number of vertices on each side of the inner $6 \times 6$ square, at least two of the corner vertices of this square (out of a total of 49) are not vertices of the graph. Therefore, the graph $G$ has at most $47+k$ vertices, so $47+k \geqslant 4k+1$, i.e., $k \leqslant 15$.

The image shows that the number 15 can be achieved.

![](https://cdn.mathpix.com/cropped/2024_06_04_b16abd6d0e010d19edb8g-3.jpg?height=368&width=368&top_left_y=1912&top_left_x=1415)
3. How many fields of the $8 \times 8$ chessboard can be cut along both diagonals so that the board does not fall apart?