Example 5 Fill the numbers $1,2, \cdots, 9$ into a $3 \times 3$ chessboard, such that the sum of the absolute values of the differences between adjacent (sharing a common edge) cells is maximized. Find this maximum value.
Solution: Let the sum of the absolute values of the differences between adjacent cells be $M$. Note that $M$ can be divided into two parts:
(1) The sum of the absolute values of the differences between adjacent numbers in the 8 surrounding cells, denoted as $M_{1}$;
(2) The sum of the 4 absolute values of the differences between the numbers adjacent to the center cell and the number in the center cell, denoted as $M_{2}$.
Thus, we have $M=M_{1}+M_{2}$.
From Example 4, we know that when the numbers in the 8 surrounding cells are rotated, the value of $M_{1}$ remains unchanged, and at this time, $M_{2}$ has only two different values, hence, $M$ also has only two different values. Let the numbers in the cells of the chessboard be $a_{1}, a_{2}, \cdots, a_{9}$, with $a_{9}$ in the center cell, and the rest $a_{i}(i=1,2, \cdots, 8)$ in the 8 surrounding cells in sequence.

From Example 4, we also know that when $a_{9}$ is fixed, $a_{1}, a_{2}, \cdots, a_{8}$ are arranged in an alternating order of size, $M_{1}$ is maximized.

If $6 \leqslant a_{9} \leqslant 9$, when the 4 larger numbers are placed in the four corner cells of the $3 \times 3$ chessboard, $M_{2}$ is maximized;

If $1 \leqslant a_{9} \leqslant 5$, when the 4 smaller numbers are placed in the four corner cells of the $3 \times 3$ chessboard, $M_{2}$ is maximized.
Below, we use the enumeration method to analyze:
When $a_{9}=9$,
$M_{1} \leqslant 2 \times(8+7+6+5)-2 \times(4+3+2+1)=32$,
$M_{2} \leqslant(9-1)+(9-2)+(9-3)+(9-4)=26$,
Thus, $M=M_{1}+M_{2} \leqslant 32+26=58$;
When $a_{9}=1$,
$M_{1} \leqslant 2 \times(9+8+7+6)-2 \times(5+4+3+2)=32$,
$M_{2} \leqslant(9-1)+(8-1)+(7-1)+(6-1)=26$,
Thus, $M=M_{1}+M_{2} \leqslant 32+26=58$;
Similarly,
When $a_{9}=8$ or 2, $M_{1} \leqslant 34, M_{2} \leqslant 22$, then
$M \leqslant 56$;
When $a_{9}=7$ or 3, $M_{1} \leqslant 36, M_{2} \leqslant 18$, then
$M \leqslant 54$;
When $a_{9}=6$ or 4, $M_{1} \leqslant 38, M_{2} \leqslant 14$, then
$M \leqslant 52$;
When $a_{9}=5$, $M_{1} \leqslant 40, M_{2} \leqslant 10$, then
$M \leqslant 50$.
In summary, the $3 \times 3$ chessboard filled as shown in Figure 8, makes $M$ have the maximum value, and the maximum value is 58.