1. In the black cells of an $8 \times 8$ chessboard, the numbers $1,2,3, \ldots, 32$ are placed such that the sum of the numbers in any $2 \times 2$ square does not exceed $S$. Find the smallest possible value of $S$.
Answer: 33.

Solution. Divide the chessboard into 16 squares of $2 \times 2$. Since the sum of the numbers in all the black cells is

$$
1+2+\cdots+32=\frac{32 \cdot 33}{2}=16 \cdot 33
$$

the arithmetic mean of the sums of the numbers in these 16 squares is 33. Therefore, in at least one square, the sum of the numbers is not less than 33, i.e., $S \geqslant 33$. An example of an arrangement where $S=33$ is achieved is shown in the figure.

| 32 |  | 31 |  | 30 |  | 29 |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | 1 |  | 2 |  | 3 |  | 4 |
| 28 |  | 27 |  | 26 |  | 25 |  |
|  | 5 |  | 6 |  | 7 |  | 8 |
| 24 |  | 23 |  | 22 |  | 21 |  |
|  | 9 |  | 10 |  | 11 |  | 12 |
| 20 |  | 19 |  | 18 |  | 17 |  |
|  | 13 |  | 14 |  | 15 |  | 16 |