2.91 If in set $A$, there exist $m$ elements $a_{1}, a_{2}, \cdots, a_{m}$ (which can be repeated), satisfying the equation $a_{1}+a_{2}+\cdots+a_{m-1}=a_{m}$, we say that set $A$ is $m$-sum-related. Given that the set $\left\{1,2, \cdots, n_{m}\right\}$ is arbitrarily divided into the union of two disjoint subsets, and at least one of these subsets is $m$-sum-related $(m \geqslant 3)$, find the minimum value of the natural number $n_{m}$.
[Solution] For $\left\{1,2, \cdots, m^{2}-m-2\right\}$, we decompose it as follows:
$$
\begin{array}{l}
S=\left\{1,2, \cdots, m-2,(m-1)^{2}, \cdots, m^{2}-m-2\right\}, \\
T=\left\{m-1, m, \cdots, m^{2}-2 m\right\} .
\end{array}
$$

At this point, the sum of any $m-1$ terms in $T$ is not less than $(m-1)^{2}>m^{2}-2 m$, so $T$ is not $m$-sum-related. If $m-1$ terms in $S$ do not contain $(m-1)^{2}$ and subsequent elements, then their sum does not exceed $(m-1)(m-2)m^{2}-m-2$, so $S$ is also not $m$-sum-related. This indicates that the smallest positive integer $n_{m} \geqslant m^{2}-m-1$.

When $n=m^{2}-m-1$, let $\{1,2, \cdots, n\}=S \cup T$, and $S \cap T=\varnothing$. Assume $1 \in S$. If $S$ and $T$ are not $m$-sum-related, then it must be that $1 \cdot(m-1)=m-1 \in T$. Thus, $(m-1)^{2} \in S$. Since $1+(m-2) m=(m-1)^{2}$, it follows that $m \in T$. This means $m, m-1 \in T, 1, (m-1)^{2} \in S$, and
$$
(m-2) \cdot m+1 \cdot(m-1)=m^{2}-m-1=(m-2) \cdot 1+1 \cdot(m-1)^{2},
$$
which implies that whether $m^{2}-m-1$ belongs to $S$ or $T$, it contradicts the assumption of the proof by contradiction. This indicates that at least one of $S$ and $T$ is $m$-sum-related.
In summary, the minimum value of $n_{m}$ is $m^{2}-m-1$.