$2 \cdot 87$ Try to find the smallest natural number $n$, such that when the set $\{1,2, \cdots, n\}$ is arbitrarily divided into two non-intersecting subsets, it is always possible to select 3 different numbers from one of the subsets, where the product of two of the numbers equals the third number.
[Solution 1] Following the principle of grouping from smallest to largest, the first 95 natural numbers can be divided into two groups: $\square$
$$
\begin{aligned}
A= & \{1,2,3,4,5,7,9,11,13,48,60,72,80,84,90\}, \\
B= & \{6,8,10,12,14,15, \cdots, 47,49,50, \cdots, 59,61, \cdots, 71,73, \cdots, 79,81, \\
& 82,83,85, \cdots, 89,91, \cdots, 95\} .
\end{aligned}
$$

It is easy to see that in both $A$ and $B$, no three numbers satisfy the problem's requirement. However, when 96 is added to either $A$ or $B$, three numbers will satisfy the problem's requirement $(2 \times 48=96=8 \times 12)$.

Below, we use proof by contradiction to show that $n=96$ satisfies the problem's requirement. Suppose $\{1,2, \cdots, 96\}=A \cup B$ and there are no three numbers that satisfy the requirement. Since $2 \times 48=3 \times 32=96$, if 2, 48 $\in A$, 3, 32 $\in B$ or vice versa, then whether 96 belongs to $A$ or $B$, three numbers will satisfy the requirement. Therefore, these pairs must either all be in one set or at least one pair must be split.
(1) If $2,3,32 \in A$, then because $2 \times 3=6, 32 \div 2=16, 3 \times 32=96$, it follows that $6,16,96$ must all belong to $B$, which is a contradiction.
(2) If $2,3,48 \in A$, then because $2 \times 3=6, 48 \div 3=16, 2 \times 48=96$, it again leads to $6,16,96$ all belonging to $B$, which is a contradiction.
(3) If $2,3 \in A, 32,48 \in B$, then $6 \in B, 8 \in A$. Thus, $4 \in B, 24 \in B$, which is a contradiction.
(4) If $2,32 \in A, 3,48 \in B$, then whether 16 belongs to $A$ or $B$, it leads to a contradiction.
(5) If $2,32,48 \in A, 3 \in B$, then $16,24,96 \in B$. Thus, $8 \in A, 6 \in B$. But $6 \times 16=96$, which is a contradiction.
(6) If $3,32,48 \in A, 2 \in B$, then $16,96 \in B$. Thus, $8 \in A, 6 \in B$, which is a contradiction.
In summary, the smallest natural number $n=96$.
[Solution 2] Using the example from Solution 1, we know that the smallest natural number $n \geqslant 96$. Below, we prove that for any decomposition $\{1,2, \cdots, 96\}=A \cup B$, there is always a set $A$ or $B$ that contains three numbers satisfying the problem's requirement.

If not, assume without loss of generality that $2 \in A$. Since $2 \times 3=6, 2 \times 4=8$, at least one of 3 and 6, and 4 and 8 must be in $B$.
(1) If $3 \in B, 4 \in B$. Since $3 \times 4=12$, it follows that $12 \in A$. Also, since $2 \times 6=12, 2 \times 12=24$, it follows that $6 \in B, 24 \in B$. But $4 \times 6=24$ cannot both be in $B$, which is a contradiction.
(2) If $3 \in B, 8 \in B$. Since $3 \times 8=24$, it follows that $24 \in A$. Also, since $2 \times 12=24, 2 \times 24=48$, it follows that $12 \in B, 48 \in B$. Since $3 \times 4=12, 6 \times 8=48$, it follows that $4 \in A, 6 \in A$, i.e., $\{4,6,24\} \subset A$, which is a contradiction.
(3) If $6 \in B, 4 \in B$. Then $24 \in A$. Therefore, $12 \in B, 48 \in B$. Since $3 \times 4=12, 6 \times 8=48$, it follows that $3 \in A, 8 \in A$, i.e., $\{3,8,24\} \subset A$, which is a contradiction.
(4) If $6 \in B, 8 \in B$. Then $48 \in A$. Therefore, $24 \in B, 96 \in B$. Since $3 \times 8=24, 6 \times 16=96$, it follows that $3 \in A, 16 \in A$, i.e., $\{3,16,48\} \subset A$, which is a contradiction. In summary, the smallest natural number $n=96$.