Example 2 Let $U=\{1,2,3, \cdots, n\}(n \geqslant 3), S$ be a subset of $U$. If an element not in $S$ appears somewhere in a permutation $P$ of $U$, and in this permutation $P$ it is between two elements of $S$, then we say $S$ is split by the permutation $P$ (for example, the permutation 13542 can split the subset $\{1,2,3\}$, but cannot split the subset $\{3,4,5\}$). Try to find the maximum positive integer $m$, such that for any $m$ subsets of $U$ (each subset has at least 2 elements and at most $n-1$ elements), there exists a permutation of $U$ that splits all these $m$ subsets.
Solve for $n=3$, there does not exist a permutation of $U=\{1,2,3\}$ that splits into two subsets $\{1,2\}$ and $\{1,3\}$. However, for any 2-element subset $A$ of $\{1,2,3\}$, there always exists a permutation of $U$ that splits $A$. That is, when $n=3$, the maximum value of $m$ is 1. Similarly, it can be deduced that when $n=4$, the maximum value of $m$ is 2. Generally, we conjecture that for any positive integer $n \geqslant 3$, the maximum value of $m$ is $n-2$.

First, we prove by mathematical induction: For $U=\{1,2, \cdots, n\}(n \geqslant 3)$, any $n-2$ subsets (each subset has at least 2 elements and at most $n-1$ elements), there must exist a permutation $P$ of $U$ that splits these $n-2$ subsets.

When $n=3$, for $U=\{1,2,3\}$, the $n-2=1$ two-element subset $\{i, j\}$ can be split by the permutation $(i, k, j)$ of $U$, where $k$ is the third element of $U$ other than $i, j$. Therefore, the conclusion holds for $n=3$.

Assume the conclusion holds for $n \geqslant 3$. Let $U=\{1,2,3, \cdots, n, n+1\}$, and $\mathscr{J}$ be a family of $n-1$ subsets of $U$, where each subset has at least 2 elements and at most $n$ elements.

Let $\mathscr{J}$ contain $k$ two-element subsets and $l$ $n$-element subsets, $k+l \leqslant n-1$. Since the $k$ two-element subsets contain $2k$ elements (including repeated counts), there are at most $k$ elements that appear at least twice, and thus at least $n+1-k \geqslant n+1-(n-1-l)=l+2$ elements appear at most once. If $l=0$, then $\mathscr{J}$ has no $n$-element subsets; if $l \geqslant 1$, then there are only $l$ elements in $U$ that do not belong to these $l$ $n$-element subsets (since for each $n$-element subset, there is exactly one element in $U$ that does not belong to it), so among these $l+2$ elements, there must be one element that belongs to all $l$ $n$-element subsets. Thus, we have proved the following conclusion: there is an element $a$ in $U$ that belongs to at most one two-element subset and to all $l$ $n$-element subsets, or $\mathscr{J}$ has no $n$-element subsets.

Assume $a=n+1$, otherwise, we can rearrange the elements of $U$. Removing $a=n+1$, all $n$-element subsets in $\mathscr{J}$ become $(n-1)$-element subsets of $\{1,2, \cdots, n\}$ (or $\mathscr{J}$ originally had no $n$-element subsets), and at most one two-element subset becomes a single-element set.

If there is a two-element subset that becomes a single-element set $\{i\}$, by the induction hypothesis, there exists a permutation $P$ of $\{1,2, \cdots, n\}$ that splits all other remaining $n-2$ subsets. Adding $a=n+1$ to the permutation $P$ such that $n+1$ is not adjacent to $i$, we get a permutation $P^{\prime}$ of $U=\{1,2, \cdots, n, n+1\}$ that splits all $n-1$ subsets in $\mathscr{J}$.

If there is no such single-element set, then choose any one of the $n-1$ subsets $S$. By the induction hypothesis, there exists a permutation $P$ of $\{1,2, \cdots, n\}$ that splits the remaining $n-2$ subsets. If $n+1 \notin S$, then adding $a=n+1$ to $P$ to get the permutation $P^{\prime}$ and placing $n+1$ between the two elements of $S$ in $P^{\prime}$, $P^{\prime}$ splits all $n-1$ subsets in $\mathscr{J}$. If $n+1 \in S$, then adding $n+1$ to the left or right end of $P$ to get the permutation $P^{\prime}$, $P^{\prime}$ splits $S$, thus $P^{\prime}$ splits all $n-1$ subsets in $\mathscr{J}$.

Thus, we have proved that for $n \geqslant 3$, the stated conclusion holds.
On the other hand, for any $n \geqslant 3$, there does not exist any permutation of $U=\{1,2, \cdots, n\}$ that splits the following $n-1$ subsets of $U$:
$\{1,2\},\{1,3\},\{1,4\}, \cdots,\{1, n\}$.
In fact, the permutations that split the subsets $\{1,2\},\{1,3\}, \cdots,\{1, n-1\}$ can only have the following forms: $P_{1}=\left(1 n a_{2} a_{3} \cdots a_{n-1}\right)$ or $P_{2}=\left(a_{2} a_{3} \cdots a_{n-1} n 1\right)$, where $\left(a_{2} a_{3} \cdots a_{n-1}\right)$ is any permutation of $(2,3,4, \cdots,(n-1))$, but neither $P_{1}$ nor $P_{2}$ splits the subset $\{1, n\}$.
In summary, the maximum value of $m$ is $n-2$.