$2 \cdot 72$ 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 $(a+b) \mid a b$.
[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 $(a_{1}, b_{1})=1, a_{1} \neq b_{1}$, without loss of generality, assume $a_{1}>b_{1}$.
Since
$$
\begin{array}{l}
a+b=c\left(a_{1}+b_{1}\right), \\
a b=c^{2} a_{1} b_{1},
\end{array}
$$

Therefore,
$$
\left(a_{1}+b_{1}\right) \mid c a_{1} b_{1} \text {. }
$$

Also, since $\left(a_{1}+b_{1}, a_{1}\right)=1$,
$$
\left(a_{1}+b_{1}, b_{1}\right)=1,
$$

Therefore,
$$
\left(a_{1}+b_{1}\right) \mid c \text {. }
$$

Thus $a, b \in S$, so $a+b \leqslant 99$, i.e., $c\left(a_{1}+b_{1}\right) \leqslant 99$, so $3 \leqslant$ $a_{1}+b_{1} \leqslant 9$.

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, as listed below:
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 $a_{1}+b_{1}=4$, 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 are
$(30,6)$;
When $a_{1}+b_{1}=7$, there are
$(42,7),(35,14),(28,21)$;
When $a_{1}+b_{1}=8$, there are
$(40,24)$;
When $a_{1}+b_{1}=9$, there are
$(45,36)$.
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
$$
T=S-M,
$$

then any two numbers in $T$ cannot form a pair $(a, b)$ that meets the requirement. Since $|M|=12$, $|T|=50-12=38$, the smallest natural number $K \geqslant 39$.
On the other hand, the following 12 pairs that meet the requirement of the problem 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 $K=39$.