1. In the cells of a $5 \times 5$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table.
Answer: 48.

Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 5. Since all these sums are distinct, the minimal possible set of their values is $\{5,6, \ldots, 13,14\}$. By adding the sums of the rows and columns of the table, we get twice the sum $S$ of all the numbers in the table, as each of them is counted twice - in the row and in the column. Then

$$
S \geqslant \frac{1}{2}(5+6+\ldots+13+14)=\frac{1}{2} \cdot 95=47 \frac{1}{2}, \quad \text { that is } \quad S \geqslant 48 .
$$

An example for $S=48$ is given below.

| 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- |
| 1 | 1 | 1 | 2 | 2 |
| 1 | 1 | 2 | 3 | 3 |
| 1 | 2 | 2 | 3 | 3 |
| 2 | 3 | 3 | 3 | 4 |