Example 5 Use $0,1, \cdots, 9$ to form a ten-digit number without repeated digits that can be divided by 11. Find the largest and smallest numbers among them. ${ }^{[1]}$
Let the ten-digit number be $A=\overline{x_{1} x_{2} \cdots x_{10}}$.
$$
\begin{array}{l}
\text { Let } x=x_{1}+x_{3}+x_{5}+x_{7}+x_{9}, \\
y=x_{2}+x_{4}+x_{6}+x_{8}+x_{10} .
\end{array}
$$

Then 11 । $(x-y)$.
Now we estimate the range of $x-y$.
$$
\begin{array}{l}
\text { By } 0+1+\cdots+4 \leqslant x, y \leqslant 5+6+\cdots+9 \\
\Rightarrow 10 \leqslant x, y \leqslant 35 .
\end{array}
$$

Thus, $|x-y|=0,11,22$.
Since $x_{1}, x_{2}, \cdots, x_{10}$ is a permutation of $0,1, \cdots, 9$, we have,
$$
x_{1}+x_{2}+\cdots+x_{10}=45 \text {. }
$$

And $x-y$ has the same parity as $x+y$, so $x-y$ is odd.
Therefore, $x-y= \pm 11$.
Solving, we get $(x, y)=(28,17)$ or $(17,28)$.
(1) Find the maximum value.

If the first six digits are all the largest, i.e.,
$$
A=\overline{987654 x_{7} x_{8} x_{9} x_{10}} \text {, }
$$

then $x=9+7+5+x_{7}+x_{9}>17$,
$$
y=8+6+4+x_{8}+x_{10}>17 \text {. }
$$

This contradicts $(x, y)=(28,17)$ or $(17,28)$.
Therefore, at most the first five digits can be the largest numbers.
Let $A=\overline{98765 x_{6} x_{7} x_{8} x_{9} x_{10}}$. Then
$$
\begin{array}{l}
x=9+7+5+x_{7}+x_{9} \geqslant 21+1+0>17 \\
\Rightarrow x=28, y=17 \\
\Rightarrow x_{7}+x_{9}=7, x_{6}+x_{8}+x_{10}=3 .
\end{array}
$$

But $x_{6}+x_{8}+x_{10} \geqslant 0+1+2=3$, so $x_{6}, x_{8}, x_{10}$ is a permutation of $0, 1, 2$.
Thus, $x_{7}, x_{9}$ is a permutation of 3, 4, and in this case, $A \leqslant 9876524130$.
If $\overline{x_{1} x_{2} x_{3} x_{4}} \neq 9876$, then $A \leqslant 98759999991024375869$.
In summary, $A_{\min }=1024375869$.