1. In the cells of a $10 \times 10$ table, the numbers $1,2,3, \ldots, 100$ are arranged 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: 202.

Solution. Divide the $10 \times 10$ table into 25 squares of $2 \times 2$. Since the sum of the numbers in the entire table is

$$
1+2+\cdots+100=\frac{100 \cdot 101}{2}=5050
$$

the arithmetic mean of the sums of the numbers in these 25 squares is 202. Therefore, in at least one square, the sum of the numbers is not less than 202, that is, $S \geqslant 202$. An example of an arrangement where the value $S=202$ is achieved is shown in the figure.

| 100 | 99 | 98 | 97 | 96 | 95 | 94 | 93 | 92 | 91 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 90 | 89 | 88 | 87 | 86 | 85 | 84 | 83 | 82 | 81 |
| 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| 80 | 79 | 78 | 77 | 76 | 75 | 74 | 73 | 72 | 71 |
| 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
| 70 | 69 | 68 | 67 | 66 | 65 | 64 | 63 | 62 | 61 |
| 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |
| 60 | 59 | 58 | 57 | 56 | 55 | 54 | 53 | 52 | 51 |
| 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |