54. (Mongolia 4) Find the smallest natural number $\boldsymbol{n}$ such that if the set $\{1,2, \cdots, n\}$ is arbitrarily divided into two non-intersecting subsets, then one of the subsets contains 3 different numbers, where the sum of two of them equals the third.
20. (Mongolia 4)

Solution: The required minimum value is 96.
Let $A_{\mathrm{s}}=\{1,2, \cdots, n\}$ be divided into two subsets $B_{\mathrm{n}}$ and $C_{\mathrm{s}}$, and neither $B_{\mathrm{s}}$ nor $C_{\mathrm{s}}$ contains three distinct numbers where the product of two equals the third. If $n \geqslant 96$, then the divisors of 96, $\{1,2,3,4,6,8,12,16,24,32,48,96\}$, cannot be separated without contradiction. Assume $1,2 \subseteq B_{0}$. This leads to four cases:
$$
\begin{array}{l}
A: 3 \in B_{0}, 4 \in B_{0} ; \\
B: 3 \in B_{0}, 4 \in C_{0} ; \\
C: 3 \in C_{0}, 4 \in B_{0} ; \\
D: 3 \in C_{0}, 4 \in C_{1} .
\end{array}
$$

In case $A$, $2,3,4 \in B_{0}$, so 6, 8, and $k 2 \in C_{0}$. Also, $48,96 \in B_{\mathrm{n}}$, and from $2,48,96 \in B$, a contradiction arises. Similarly, in case $B$, $2,3 \in B_{a}, 4 \in C_{\mathrm{a}}$, leading to $6 \in C_{\mathrm{s}}$, and thus $24 \in B_{\mathrm{s}}$. Consequently, $8,12,48 \in C_{0}$, and since $6 \cdot 8=48$, a contradiction arises. In case $C$, $2,4 \in B_{\mathrm{a}}, 3 \in C_{\mathrm{a}}$, so $8 \in C_{\mathrm{s}}, 24 \in B_{\mathbb{\pi}}$, and thus $6,48 \in C_{\mathrm{a}}$. Again, $6 \cdot 8=48$ leads to a contradiction. Finally, in case $D$, $2 \in B_{\text {: }}$, $3,4 \in C_{\mathrm{s}}$, so $12 \in B_{\mathrm{a}}$, and $6,24 \in C_{\mathrm{a}}$, and since $4 \cdot 6=24$, a contradiction arises.

For example, $B_{\mathrm{a}}=\left\{1, p, p^{2}, p^{2} q r, p^{4} q, p^{3} q^{2}\right\}$
and $C_{0}=\left\{p^{3}, p^{4}, p^{5}, p^{8}, p q, p^{2} q, p q r, p^{3} q \cdot p^{2} q^{2}\right\}$, where $p, q, r$ are different primes. That is, $B_{8 \overline{0}}=\{1,2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41,43,47,48,49,53,59,60,61,67,71,72,73,79,80,83,84,89,90\}$.