Example 6 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 $(a+b) \mid a b$.
(1993 China Winter Camp Problem)
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 $a+b=c(a_{1}+b_{1}), a b=c^{2} a_{1} b_{1}$,
it follows that $(a_{1}+b_{1}) \mid c a_{1} b_{1}$.
Also, since $(a_{1}+b_{1}, a_{1})=1, (a_{1}+b_{1}, b_{1})=1$,
it follows that $(a_{1}+b_{1}) \mid c$.
Thus, $a, b \in S$, so $a+b \leqslant 99$, i.e.,
$c(a_{1}+a_{1}) \leqslant 99$, which means $3 \leqslant a_{1}+b_{1} \leqslant 9$.
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, listed as follows:
When $a_{1}+b_{1}=3$, $c$ is a multiple of 3, and 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 $a_{1}+b_{1}=4$, there are $(12,4),(24,8),(36,12),(48,16)$;
When $a_{1}+b_{1}=5$, there are $(20,5),(40,10),(15,10),(30,20),(45,30)$;
When $a_{1}+b_{1}=6$, there is $(30,6)$;
When $a_{1}+b_{1}=7$, there are $(42,7),(35,14),(28,21)$;
When $a_{1}+b_{1}=8$, there is $(40,24)$;
When $a_{1}+b_{1}=9$, there is $(45,36)$.
Let $M=\{6,12,15,18,20,21,24,35,40,42,45,48\}$, then each of the 23 pairs above contains at least one element from $M$. Let $T=S-M$.

Thus, any two numbers in $T$ cannot form a pair $(a, b)$ that meets the requirement. Since $|M|=12, |T|=50-12=38$, it follows that the smallest natural number $k \geqslant 39$.
On the other hand, the following 12 pairs that satisfy the problem's requirements are 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 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 $k=39$.