20. (MON 4) Find the least natural number $n$ such that if the set $\{1,2, \ldots, n\}$ is arbitrarily divided into two nonintersecting subsets, then one of the subsets contains three distinct numbers such that the product of two of them equals the third.
20. Suppose that $A_{n}=\{1,2, \ldots, n\}$ is partitioned into $B_{n}$ and $C_{n}$, and that neither $B_{n}$ nor $C_{n}$ contains 3 distinct numbers one of which is equal to the product of the other two. If $n \geq 96$, then the divisors of 96 must be split up. Let w.l.o.g. $2 \in B_{n}$. There are four cases.
(i) $3 \in B_{n}, 4 \in B_{n}$. Then $6,8,12 \in C_{n} \Rightarrow 48,96 \in B_{n}$. A contradiction for $96=2 \cdot 48$.
(ii) $3 \in B_{n}, 4 \in C_{n}$. Then $6 \in C_{n}, 24 \in B_{n}, 8,12,48 \in C_{n}$. A contradiction for $48=6 \cdot 8$.
(iii) $3 \in C_{n}, 4 \in B_{n}$. Then $8 \in C_{n}, 24 \in B_{n}, 6,48 \in C_{n}$. A contradiction for $48=6 \cdot 8$.
(iv) $3 \in C_{n}, 4 \in C_{n}$. Then $12 \in B_{n}, 6,24 \in C_{n}$. A contradiction for $24=4 \cdot 6$.
If $n=95$, there is a very large number of ways of partitioning $A_{n}$. For example, $B_{n}=\left\{1, p, p^{2}, p^{3} q^{2}, p^{4} q, p^{2} q r \mid p, q, r=\right.$ distinct primes $\}$, $C_{n}=\left\{p^{3}, p^{4}, p^{3}, p^{6}, p q, p^{2} q, p^{3} q, p^{2} q^{2}, p q r \mid p, q, r=\right.$ distinct primes $\}$. Then $B_{95}=\{1,2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41$, $43,47,48,49,53,59,60,61,67,71,72,73,79,80,83,84,89,90\}$.