7.80 Given that each of the $h$ chessboard pieces (from international chess) has 64 squares numbered from 1 to 64, such that when any two chessboard pieces are placed with their perimeters overlapping in any way, the numbers on any two overlapping squares are never the same, find the maximum value of the number of chessboard pieces $h$.


[Solution] Divide the 64 squares on the chessboard into 16 groups as shown in the figure on the right. Divide the 64 natural numbers from 1 to 64 into 16 groups, with 4 numbers in each group. Fill these 16 groups of numbers into the $A, B, C, \cdots, P$ groups of squares on the first chessboard. For the second chessboard, fill the 16 groups of numbers into the $B, C, \cdots, P, A$ groups of squares. Continue this rotation, and finally fill the 16 groups of numbers into the $P, A, B, \cdots, O$ groups of squares on the 16th chessboard. It is easy to see that the numbering of these 16 chessboards meets the requirements of the problem, so the maximum value of $h$ is at least 16.

If there are 17 chessboards that have been numbered, then the $A$ group squares on these chessboards have a total of 68 numbers. By the pigeonhole principle, there must be two identical numbers among them, and these two identical numbers are not on the same chessboard. Thus, it is possible to choose the positions of these two chessboards so that the two $A$ group squares coincide. Therefore, any numbering method for 17 chessboards does not meet the requirements. Hence, the maximum value of $h$ is 16.