Example 4 Given a positive integer $n \geqslant 3$, find the smallest positive integer $m$ with the following property: if the set $S=\{1,2, \cdots, m\}$ is arbitrarily divided into the union of two non-empty, disjoint subsets, then one of these subsets must contain $n$ numbers (not necessarily distinct) $x_{1}, x_{2}, \cdots, x_{n}$, such that $x_{1}+x_{2}+\cdots+x_{n-1}=x_{n}$.
We first estimate the minimum value of $m$ for which the set $S=\{1,2, \cdots, m\}$ has the property described in the problem. If $S$ does not have the property, then there exist two non-empty, disjoint subsets $A$ and $B$ of $S$ such that $S=A \cup B$, and the sum of any $n-1$ numbers (not necessarily distinct) in $A(B)$ is not in $A(B)$.

Assume $1 \in A$, then $\underbrace{1+1+\cdots+1}_{n-1 \uparrow}=n-1 \in B$ (as long as $m \geqslant n-1$), thus $\underbrace{(n-1)+(n-1)+\cdots+(n-1)}=(n-1)^{2} \in A$ (as long as $m \geqslant(n-1)^{2})$.
If $n \in A$, then $(n-1)^{2}=\underbrace{n+n+\cdots+n}_{n-2 \uparrow}+1 \in B$, which is a contradiction.
If $n \in B$, then $\underbrace{n+n+\cdots+n}_{n-2 \uparrow}+(n-1)=n^{2}-n-1 \in A$ (as long as $m \geqslant n^{2}-n-1$), but $\underbrace{1+1+\cdots+1}_{n-2 \uparrow}+(n-1)^{2}=n^{2}-n-1 \in B$, which is a contradiction.
Therefore, when $m \geqslant n^{2}-n-1$, the set $S=\{1,2, \cdots, m\}$ has the property described in the problem.
Next, when $m \leqslant n^{2}-n-2$, for the set $S=\{1,2,3, \cdots, m\}$, let $A=\{1,2, \cdots, n-2, (n-1)^{2},(n-1)^{2}+1, \cdots, m\}, B=\left\{(n-1), n, \cdots,(n-1)^{2}-1\right\}$, then $A \cap B=\varnothing, S=A \cup B$. We prove that there do not exist $x_{1}, x_{2}, \cdots, x_{n-1} \in A$ (or $B$) such that $x_{1}+x_{2}+\cdots+x_{n-1} \in A$ (or $B$) (here $x_{1}, x_{2}, \cdots, x_{n-1}$ are not necessarily distinct).

In fact, if $x_{1}, x_{2}, \cdots, x_{n-1} \in\{1,2, \cdots, n-2\} \subset A$, then $n-1 \leqslant x_{1}+x_{2}+\cdots+x_{n-1} \leqslant (n-1)(n-2)<(n-1)^{2}$, hence $x_{1}+x_{2}+\cdots+x_{n-1} \notin A$.

If $x_{1}, x_{2}, \cdots, x_{n-1}$ contains at least one element from $\left\{(n-1)^{2},(n-1)^{2}+1, \cdots, n^{2}-n-2\right\}$, then $x_{1}+x_{2}+\cdots+x_{n-1} \geqslant \underbrace{1+1+\cdots+1}_{n-2 \uparrow}+(n-1)^{2}=n^{2}-n-1$, hence $x_{1}+x_{2}+\cdots+x_{n-1} \notin A$.
If $x_{1}, x_{2}, \cdots, x_{n-1} \in B$, then $x_{1}+x_{2}+\cdots+x_{n-1} \geqslant(n-1)^{2}$, hence $x_{1}+x_{2}+\cdots+x_{n-1} \notin B$.
Therefore, when $m \leqslant n^{2}-n-2$, the set $S=\{1,2, \cdots, m\}$ does not have the property described in the problem.
In summary, the minimum value of $m$ is $n^{2}-n-1$.