1. In each cell of an $n \times n$ table, a natural number is written. It turns out that for all $k$ from 1 to $n$, the sum of the numbers in the $k$-th column differs by one from the sum of the numbers in the $k$-th row. For which $n$ is this possible?
Answer: for even $n$

Solution. If $n$ is odd, then in the table, the sum of all numbers, calculated by rows, differs from the sum of numbers, calculated by columns, by an odd number. But this is impossible. In a table with even $n$, the upper half of the diagonal running from the bottom left corner to the top right corner can be filled with twos, and the remaining cells can be filled with ones. The described arrangement of numbers for $n=10$ is shown in the figure.

| ![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-26.jpg?height=430&width=435&top_left_y=800&top_left_x=216) |
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