$2 \cdot 105$ Let $k, m, n$ be integers, and satisfy $1<n \leqslant m-1 \leqslant k$. Try to find the maximum order of the subset $s$ of the set $\{1,2, \cdots, k\}$, such that in $s$, the sum of any $n$ different elements is not equal to $m$.
[Solution] Let $mm \\
\end{array}
$$

be the smallest natural number. Then the sum of any $n$ different elements in the set $\{l+1, l+2, \cdots, k\}$ is greater than $m$. Also, from
$$
n l+\frac{n(n+1)}{2}>m
$$

we get
$$
\begin{array}{l}
l>\frac{m}{n}-\frac{n+1}{2}, \\
l=\left[\frac{m}{n}-\frac{n+1}{2}\right]+1=\left[\frac{m}{n}-\frac{n-1}{2}\right]
\end{array}
$$

Thus, the maximum order of $s$ is $\geqslant k-l=k-\left[\frac{m}{n}-\frac{n-1}{2}\right]$.
Next, we prove: When $m \geqslant \frac{n(n+1)}{2}$, for any $s \subseteq\{1,2, \cdots, k\}$, if $|s|>k-l$, then there exist $n$ different numbers in $s$ whose sum equals $m$.
We use proof by contradiction.
Assume there exists $s \subseteq\{1,2, \cdots, k\},|s|>k-l$, such that the sum of any $n$ different elements in $s$ is not equal to $m$. Let
$$
\begin{array}{l}
X=\left\{A|A \subset\{1,2, \cdots k\},| A \mid=n, \sum_{i \in A} i=m\right\}, \\
X_{j}=\left\{A|A \subset\{1,2, \cdots k\},| A \mid=n, j \in A, \sum_{i \in A} i=m\right\} .
\end{array}
$$

Let $t=k-|s|$, then $tn t+n-1 \\
& =(n-1) t+t+n-1,
\end{aligned}
$$

Therefore, we know
$$
m-(n-1) t-\frac{n(n-1)}{2}>t+n-1,
$$

Thus, $t+1, t+2, \cdots, t+n-1, m-(n-1) t-\frac{n(n-1)}{2}$ are $n$ different numbers in the set $\{t+1, t+2, \cdots, k\}$, and
$$
(t+1)+(t+2)+\cdots+(t+n-1)+\left[m-(n-1) t-\frac{n(n-1)}{2}\right]
$$
.$=m$,
which is a contradiction.
Hence, when $m \geqslant \frac{n(n+1)}{2}$, the maximum order of set $S$ is $k-\left[\frac{m}{n}-\frac{n-1}{2}\right]$. In summary, we have
$$
S \text { 's maximum order }=\left\{\begin{array}{l}
k, \text { when } m<\frac{n(n+1)}{2} \text {, } \\
k-\left[\frac{m}{n}-\frac{n-1}{2}\right], \text { when } m \geqslant \frac{n(n+1)}{2} \text { . }
\end{array}\right.
$$