Example 12 As shown in Figure $4-2, x_{1}, x_{2}, \cdots, x_{10}$ is a permutation of $0,1,2, \cdots, 9$, such that the sum of the numbers in each of the 4 circles is equal, denoted as $S$. Find the minimum and maximum values of $S$.

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Solve for the minimum value first.
$$
S=x_{2}+x_{3}+x_{4}+x_{5}+x_{7}+x_{8}+x_{9} \geqslant 0+1+2+3+4+5+6=21 .
$$

On the other hand, as shown in Figure 4-3, we know that $S=21$ can be achieved, so the minimum value of $S$ is 21.
Next, solve for the maximum value. From the given conditions, we easily get
$$
\begin{array}{l}
x_{1}=x_{7}+x_{8}+x_{9}, \\
x_{6}=x_{2}+x_{5}+x_{9}, \\
x_{10}=x_{2}+x_{3}+x_{7},
\end{array}
$$

Thus, $x_{1}+x_{6}+x_{10}=2\left(x_{2}+x_{7}+x_{9}\right)+\left(x_{3}+x_{5}+x_{8}\right)$.
Also, $S+\left(x_{1}+x_{6}+x_{10}\right)=\sum_{i=1}^{10} x_{i}=0+1+2+\cdots+9=45$,
$$
\begin{array}{l} 
\text { Therefore, } 45+\left(x_{2}+x_{7}+x_{9}\right)=x_{4}+\left(x_{1}+x_{6}+x_{10}\right)+2\left(x_{2}+x_{7}+x_{9}\right)+\left(x_{3}+x_{5}+x_{8}\right) \\
=x_{4}+2\left(x_{1}+x_{6}+x_{10}\right)=x_{4}+2(45-S), \\
2 S=45+x_{4}-\left(x_{2}+x_{7}+x_{9}\right) \leqslant 45+9-(0+1+2)=51 .
\end{array}
$$

Since $S$ is a positive integer, $S \leqslant 25$.
When $S=25$, $x_{4}-\left(x_{2}+x_{7}+x_{9}\right)=5$ and $x_{1}+x_{6}+x_{10}=45-25=20$, thus $x_{4}=$ $5+\left(x_{2}+x_{7}+x_{9}\right) \geqslant 5+0+1+2=8$, so $x_{4}=8$ or $9,\left\{x_{2}, x_{7}, x_{9}\right\}=\{0,1,2\}$ or $\{0,1,3\}$. Therefore, $\left\{x_{1}, x_{6}, x_{10}\right\}=\{9,7,4\}$ or $\{8,7,5\},\left\{x_{3}, x_{5}, x_{8}\right\}=\{3,5,6\}$ or $\{2,4,6\}$. As shown in Figure 4-4 or Figure 4-5, $S=25$ can be achieved.
Therefore, the maximum value of $S$ is 25.
(5) Estimation from the opposite perspective
When it is difficult to estimate from a positive perspective using the above methods, consider estimating from the opposite perspective to derive the required maximum (minimum) value.