$1 \cdot 131$ Can $9 \times 9$ grid be filled with natural numbers 1 to 8, one number per cell, so that the sum of the 9 numbers in each $3 \times 3$ square is the same?
[Solution] First, fill out a $9 \times 9$ grid table $T$:
\begin{tabular}{c|c|c|c|c|c|c|c|c}
\hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline 3 & 4 & 5 & 6 & 7 & 8 & 0 & 1 & 2 \\
\hline 6 & 7 & 8 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline 3 & 4 & 5 & 6 & 7 & 8 & 0 & 1 & 2 \\
\hline 6 & 7 & 8 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline 3 & 4 & 5 & 6 & 7 & 8 & 0 & 1 & 2 \\
\hline 6 & 7 & 8 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
\end{tabular}

In $T$, each row contains the numbers $0,1,2, \cdots, 8$, and the sum of the 9 numbers in each $3 \times 3$ square is 36.

Rotate the grid $T$ counterclockwise by $90^{\circ}$ to get the grid $Q$. Clearly, in $Q$, the numbers $0,1,2, \cdots, 8$ each appear 9 times, and the sum of the 9 numbers in each $3 \times 3$ square is 36.

Let the number in the cell at the intersection of the $i$-th row and $j$-th column in $T$ be $a_{i j}$, and the number in the cell at the intersection of the $i$-th row and $j$-th column in $Q$ be $b_{i j}$. Then,
$$a_{i j} \neq b_{i j}, 1 \leqslant i \leqslant 9,1 \leqslant j \leqslant 9 .$$

Construct a new $9 \times 9$ grid table $P$. In $P$, the number in the cell at the intersection of the $i$-th row and $j$-th column is
$$9 a_{i j}+b_{i j}+1,1 \leqslant i \leqslant 9,1 \leqslant j \leqslant 9 .$$

Clearly, in $P$, each of the natural numbers $1,2, \cdots, 81$ appears exactly once, and the sum of the 9 numbers in each $3 \times 3$ square is
$$9 \times 36+36+9=369$$