3.87 Can natural numbers 1, 2, ..., 81 be placed in the 81 cells of a $9 \times 9$ grid, one number per cell, so that the sum of the 9 numbers in each $3 \times 3$ square is the same?
[Solution] Starting from a 3rd-order magic square and performing row cyclic shifts, we obtain the number table at the top left, where the sum of the 9 numbers filled in each $3 \times 3$ square is 45.

Rotating the number table at the top left $90^{\circ}$ counterclockwise around the center of the square, we get the number table at the top right. Clearly, the sum of the 9 numbers in each $3 \times 3$ square is also 45.
$$
\{(a, b) \mid a, b=1,2, \cdots, 9\}
$$

establishes a bijection between them.
Now, fill a new $9 \times 9$ number table, with the principle of filling the number $9a + b - 9$ in the cell corresponding to $(a, b)$. At this point, the sum of the 9 numbers in each $3 \times 3$ square is $45 \times 10 - 9 \times 9 = 369$.