At least three squares must contain more than one beetle.

Color a 9x9 checkerboard so that the diagonals share the same color, as on the left.  Notice that since movement is diagonal, a beetle must jump to a square of the same color, thus allowing you to split the problem into smaller problems: the red beetles and the black beetles.

The red beetles can all be paired up, as shown with the yellow lines, leaving no red squares with more than one beetle on it.

Now look at the black squares.  Five rows of five squares alternate with four rows of four squares. Therefore, 25 beetles must be crammed into 16 spaces.  Since at most 4 beetles can meet at a single square, the minimum number of squares where a beetle-meeting occurs is achieved when  three squares have four beetles on them, and the rest have contain a single beetle, accounting for the total of 3·4 + 13 = 25 beetles.  Such a pairing is depicted in the sketch, showing that three is the smallest number of squares that can host a beetle party.

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ã2003 Alberto L. Delgado