4. (China Mathematical Olympiad 1996) 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 $(a+b) \mid a b$.
4. Let $a, b \in S$ satisfy the condition $(a+b) \mid a b$. Let $c=(a, b)$, then $a=c a_{1}, b=c b_{1}$, where $a_{1}, b_{1} \in \mathbf{N}$ and $(a_{1}, b_{1})=1, a_{1} \neq b_{1}$, without loss of generality, assume $a_{1}>b_{1}$.
Since $a+b=c(a_{1}+b_{1}), a b=c^{2} a_{1} b_{1}$,

Therefore, $(a_{1}+b_{1}) \mid c a_{1} b_{1}$.
Also, since $(a_{1}+b_{1}, a_{1})=1, (a_{1}+b_{1}, b_{1})=1$,

Therefore, $(a_{1}+b_{1}) \mid c$.
Thus, $a, b \in S$, so $a+b \leqslant 99$, i.e., $c(a_{1}+b_{1}) \leqslant 99$, hence $3 \leqslant a_{1}+b_{1} \leqslant 9$.
From this, we know that the number of different pairs $(a, b)$ in $S$ that satisfy the condition $(a+b) \mid a b$ is 23. The pairs are listed as follows:
When $a_{1}+b_{1}=3$, $c$ is a multiple of 3, the pairs $(a, b)$ in this category are
$(6,3),(12,6),(18,9),(24,12),(30,15),(36,18),(42,21),(48,24)$;
When $a_{1}+b_{1}=4$, the pairs are $(12,4),(24,8),(36,12),(48,16)$;
When $a_{1}+b_{1}=6$, the pairs are $(30,6)$;
When $a_{1}+b_{1}=7$, the pairs are $(42,7),(35,14),(28,21)$;
When $a_{1}+b_{1}=8$, the pairs are $(40,24)$;
When $a_{1}+b_{1}=9$, the pairs are $(45,36)$.
Let $M=\{6,12,15,18,20,21,24,35,40,42,45,48\}$. Then each of the 23 pairs contains at least one element from $M$. Let
$$
T=S-M,
$$

Then any two numbers in $T$ cannot form a pair $(a, b)$ that satisfies the condition. Since $|M|=12,|T|=50-12=38$, the smallest natural number $K \geqslant 39$.
On the other hand, the following 12 pairs that satisfy the condition are mutually exclusive, they are:
$(6,3),(12,4),(20,5),(42,7),(24,8),(18,9),(40,10)$,
$(35,14),(30,15),(48,16),(28,21),(45,36)$.
For any 39-element subset $R$ of $S$, it is only 11 elements less than $S$, and these 11 elements belong to at most 11 of the above 12 pairs, so there must be one of the 12 pairs that belongs to $R$.
In summary, the smallest natural number $K=39$.