$15 \cdot 26$ Let the natural number $n(n>3)$ have the following property: when the set $S_{n}=\{3,4, \cdots, n\}$ is arbitrarily divided into two groups, there is always a group containing three numbers $a, b, c$ (allowing $a=b$), such that $a b=c$. Find the smallest value of such $n$.
(China National Training Team Test Question, 1988)
[Solution] When $n \geqslant 3^{5}$,
$$
3,3^{2}, 3^{3}, 3^{4}, 3^{5} \in S_{n} .
$$

If $S_{n}$ is divided into two groups $A$ and $B$, and neither group has the property mentioned in the problem.
Without loss of generality, assume $3 \in A$, then $3 \cdot 3 \in A$, i.e.,
$$
3^{2} \in B,
$$

Then $3^{2} \cdot 3^{2} \notin B$, i.e., $3^{4} \in A$, thus $3^{3} \notin A, 3^{3} \in B$. At this point, whether $3^{5}$ belongs to $A$ or $B$, a contradiction will arise:
If $3^{5} \in A$, then by $3 \in A, 3^{4} \in A$, a contradiction arises,
If $3^{5} \in B$, then by $3^{2} \in B, 3^{3} \in B$, a contradiction arises.
Therefore, when $n \geqslant 3^{5}=243$, the property mentioned in the problem is satisfied.
$$
\begin{aligned}
\text { If } n<3^{5} & =243, \text { let } \\
A & =\{k \mid 9 \leqslant k \leqslant 80\} \cap S_{n}, \\
B & =\{k \mid 3 \leqslant k \leqslant 8 \text { or } 81 \leqslant k \leqslant 242\} \cap S_{n},
\end{aligned}
$$

Then neither $A$ nor $B$ has the property $a b=c$.
That is, for $n<243$, there exists a grouping method such that the property mentioned in the problem is not satisfied.
Therefore, the minimum value of $n$ is 243.