# Solution

In each square of a checkered $5 \times 5$ board there is a bug. At a certain moment of time each bug moves to an adjacent square (the bug does not move diagonally). Prove that after the bugs move, there is more than one bug in one of the squares.

Without loss of generality, suppose the board is colored in the following way:

Let's argue indirectly...

Assume that, after the bugs move, there is no more than one bug in each square.

In particular, this means that there can be no more than 12 bugs on the white squares.

Notice that if the bugs moved to adjacent squares (by a side), then bugs that were on black squares are now on white squares and vice-versa. But all 13 black bugs moved to white squares, so there have to be more than 12 bugs on the white squares. We have a clear contradiction.

Consequently, our assumption must be rejected, and the opposite must be true: there is more than one bug in one of the squares.

QED.