Evaluation. Consider the numbers $a_{1}<a_{2}<\ldots<a_{8}$, standing on the black diagonal (of course, on the diagonal they may not be in order). Then $a_{1} \geq 1, a_{2} \geq 3, a_{3} \geq 5, \ldots, a_{7} \geq 13$. We will prove that $a_{8} \geq 39$.

Assume that the numbers were written into the cells of the board in ascending order. Then at the moment when the eighth number was written on the diagonal, the half of the board being left was filled (since it was not returned to). This half (including the diagonal) contains 36 cells: 16 white and 20 black. But when filling the board, white and black cells alternate. Therefore, by this moment, at least three more white cells in the other half of the board were filled, meaning a total of at least 39 cells were filled. $1+3+5+\ldots+13+39=88$

An example is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_0e910282974b56b80e3ag-36.jpg?height=578&width=575&top_left_y=1303&top_left_x=753)

## Answer

88. 

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Shapovalov A.V.

The cells of a chessboard are numbered from 1 to 64 in such a way that adjacent numbers are in adjacent (by side) cells.

What is the smallest possible sum of the numbers on the diagonal?

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