Example 4 For all real numbers $x$, such that $|x-1|+|x-2|+|x-10|+|x-11| \geqslant m$ always holds. Then the maximum value of $m$ is $\qquad$
Analysis: This type of problem is actually an extension and generalization of the function extremum problem in Example 1. It is easy to know that the function $f(x) = |x-1| + |x-2| + |x-10| + |x-11|$ has a minimum value of 18 when $x \in [2,10]$. Therefore, when $m \leqslant 18$, the inequality $f(x) \geqslant m$ always holds, i.e., the maximum value of $m$ is 18.