15.27 Find the smallest natural number $n$ with the following property: if the set $\{1,2, \cdots, n\}$ is arbitrarily divided into two disjoint subsets, then one of the subsets contains three distinct numbers, the product of two of which equals the third.
（29th International Mathematical Olympiad Candidate Problem, 1988)
[Solution] The required minimum value is 96.
Let $A_{n}=\{1,2, \cdots, n\}$. Suppose the two non-intersecting subsets are $B_{n}, C_{n}$.
First, we prove that when $n \geqslant 96$, there must be a subset containing 3 different numbers, where the product of two of them equals the third.

Assume that neither $B_{n}$ nor $C_{n}$ has this property, i.e., they do not contain such 3 different numbers where the product of two equals the third.
Since $n \geqslant 96$, the set of divisors of 96
$$
\{1,2,3,4,6,8,12,16,24,32,48,96\}
$$

cannot all be in the same subset, otherwise, there would exist two numbers whose product equals the third.
Without loss of generality, assume $2 \in B_{n}$.
For 3 and 4, there are the following four cases:
Case $A$: $3 \in B_{n}, 4 \in B_{n}$.
Case $B$: $3 \in B_{n}, 4 \in C_{n}$.
Case $C$: $3 \in C_{n}, 4 \in B_{n}$.
Case $D$: $3 \in C_{n}, 4 \in C_{n}$.
For case $A$.
Since $2,3,4 \in B_{n}$, then
$$
6 \notin B_{n}, 8 \notin B_{n}, 12 \notin B_{n}.
$$

Thus,
$$
6 \in C_{n}, 8 \in C_{n}, 12 \in C_{n}.
$$

Therefore, $48 \in B_{n}$, and since $2 \in B_{n}$, then
$$
2 \cdot 48 \in C_{n}, \text{ i.e., } 96 \in C_{n},
$$

On the other hand, $8 \in C_{n}, 12 \in C_{n}$, then
$$
8 \cdot 12 \in B_{n}, \text{ i.e., } 96 \in B_{n}.
$$

Thus, $96 \in C_{n}$ and $96 \in B_{n}$, leading to a contradiction.
Therefore, case $A$ is impossible.
For case $B$.
Since $2,3 \in B_{n}, 4 \in C_{n}$, then
$$
6 \in B_{n}, \text{ i.e., } 6 \in C_{n}.
$$

Thus,
$$
24 \in B_{n},
$$

Since $3 \in B_{n}, 24 \in B_{n}$, then $8 \in C_{n}$.
Since $2 \in B_{n}, 24 \in B_{n}$, then $12 \in C_{n}, 48 \in C_{n}$.
On the other hand, $6 \in C_{n}, 8 \in C_{n}$, then
$$
6 \cdot 8=48 \in B_{n}.
$$

From $48 \in B_{n}$ and $48 \in C_{n}$, a contradiction arises.
Therefore, case $B$ is impossible.
For case $C$,
Since $2,4 \in B_{n}, 3 \in C_{n}$, then
$$
8 \in C_{n}, \quad 24 \in B_{n}.
$$

Thus, from $4 \in B_{n}, 24 \in B_{n}$, we get $6 \in C_{n}$.
From $6 \in C_{n}, 8 \in C_{n}$, we get $48 \in B_{n}$,
From $2 \in B_{n}, 24 \in B_{n}$, we get $48 \in C_{n}$.
Thus, a contradiction arises.
For case $D$.
Since $2 \in B_{n}, 3,4 \in C_{n}$, then
$$
12 \in B_{n}, \quad 24 \in C_{n}.
$$

Since $2 \in B_{n}, 12 \in B_{n}$, then $6 \in C_{n}$.
From $4 \in C_{n}, 6 \in C_{n}$, we get $24 \in B_{n}$.
Thus, a contradiction arises from $24 \in C_{n}$ and $24 \in B_{n}$.
Therefore, case $D$ is impossible.
From the above, when $n \geqslant 96$, there must be a subset containing three numbers, where the product of two of them equals the third.
When $n<96$, a method can be constructed such that no subset contains three numbers that satisfy the condition.
$$
\begin{aligned}
\text{For } \quad A_{95}=\{1,2,3, \cdots, 95\}. \\
B_{n}=\left\{1, p, p^{2}, p^{2} q r, p^{4} q, p^{3} q^{2}\right\}, \\
C_{n}=\left\{p^{3}, p^{4}, p^{5}, p^{6}, p q, p^{2} q, p q r, p^{3} q, p^{2} q^{2}\right\}.
\end{aligned}
$$

Where $p, q, r$ are different primes not exceeding 95.
For example, $B_{95}=\{1,2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41,43$,
$47,48,49,53,59,60,61,67,71,72,73,79,80,83,84,89,90\}$.
$$
C_{95}=A_{95} \backslash B_{95}.
$$