1.92 Let $S=\{1,2, \cdots, 50\}$. Find the smallest natural number $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$, satisfying $(a+b) \mid a b$.
(11th China High School Mathematics Winter Camp, 1996)
[Solution] Let there be $a, b \in S$ satisfying the condition $(a+b) \mid a b$.
Let $c=(a, b)$, then $a=c a_{1}, b=c b_{1}$, where $a_{1}, b_{1} \in \mathbb{N}$, and $\left(a_{1}, b_{1}\right)=1$. Therefore, we have $c\left(a_{1}+b_{1}\right)=(a+b) \mid a b = c^{2} a_{1} b_{1}$,
$$
\left(a_{1}+b_{1}\right) \mid c a_{1} b_{1} \text{. }
$$

Since $\left(a_{1}, b_{1}\right)=1$, it follows that $\left(a_{1}+b_{1}, a_{1}\right)=1,\left(a_{1}+b_{1}, b_{1}\right)=1$, hence $\left(a_{1}+b_{1}\right) \mid c$.
Given $a \in S, b \in S$, then $a+b \leqslant 99$, and $c\left(a_{1}+b_{1}\right) \leqslant 99$.
Thus, $3 \leqslant a_{1}+b_{1} \leqslant 9$,
From this, we know that there are 23 different pairs in $S$ that satisfy the condition $(a+b) \mid a b$:
When $a_{1}+b_{1}=3$, $(a, b)=(6,3),(12,6),(18,9),(24,12),(30,15),(36,18),(42,21),(48,24)$;
When $a_{1}+b_{1}=4$, $(a, b)=(12,4),(24,8),(36,12),(48,16)$;
When $a_{1}+b_{1}=5$, $(a, b)=(20,5),(40,10),(15,10),(30,20),(45,30)$;
When $a_{1}+b_{1}=6$, $(a, b)=(30,6)$;
When $a_{1}+b_{1}=7$, $(a, b)=(42,7),(35,14),(28,21)$;
When $a_{1}+b_{1}=8$, $(a, b)=(40,24)$;
When $a_{1}+b_{1}=9$, $(a, b)=(45,36)$.
Let $M=\{6,12,15,18,20,21,24,35,40,42,45,48\}$, then $|M|=12$ and each of the 23 pairs above contains at least one element from $M$.

Therefore, if we let $T=S-M$, then $|T|=38$ and no two numbers in $T$ satisfy the given condition. Hence, the smallest natural number $k \geqslant 39$.
Note, the following 12 pairs that satisfy the given condition are mutually disjoint:
$(6,3),(12,4),(20,5),(42,7),(24,8),(18,9),(40,10),(35,14),(30,15),(48,16),(28,21),(45,36)$.

For any 39-element subset $R$ of $S$, it is only 11 elements less than $S$, and these 11 elements belong to at least 11 of the above 12 pairs, thus there must be one of the 12 pairs that belongs to $R$.
In summary, the smallest natural number $k=39$.