Example 1 Let $x_{1}, x_{2}, \cdots, x_{9}$ all be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots+x_{9}=220$. Then, when the value of $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is $\qquad$
$(1992$, National Junior High School Mathematics League)
Solution: From $x_{1}+x_{2}+\cdots+x_{9}=220$,
we know $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}>110$,
or $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$.
From (1), then $x_{5} \geqslant 25$. Thus, $x_{6} \geqslant 26, x_{7} \geqslant$ $27, x_{8} \geqslant 28, x_{9} \geqslant 29$. We get
$$
\begin{aligned}
\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)+\left(x_{6}+x_{7}+\right. \\
\left.x_{8}+x_{9}\right)>110+110=220 .
\end{aligned}
$$

This contradicts the given condition.
From (2), we can take $x_{1}=20, x_{2}=21, x_{3}=22$, $x_{4}=23, x_{5}=24$, so that $x_{1}+x_{2}+\cdots+x_{5}=110$ reaches its maximum value, and $x_{1}$ also reaches its maximum value. At this point, taking $x_{6}$ $=26, x_{7}=27, x_{8}=28, x_{9}=29$ satisfies all conditions. Therefore, the minimum value of $x_{9}-x_{1}$ is $29-20=9$.