2・13 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] For $a, b \in S$ satisfying $(a+b) \mid a b$, 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
$$
\begin{array}{c}
c\left(a_{1}+b_{1}\right)=(a+b) \mid a b=c^{2} a_{1} b_{1}, \\
\left(a_{1}+b_{1}\right) \mid c a_{1} b_{1} .
\end{array}
$$

Since $\left(a_{1}+b_{1}, a_{1}\right)=1,\left(a_{1}+b_{1}, b_{1}\right)=1$, it follows that
$$
\left(a_{1}+b_{1}\right) \mid c \text {. }
$$

Also, since $a, b \in S$, we have $a+b \leqslant 99$, i.e., $c\left(a_{1}+b_{1}\right) \leqslant 99$. From (1), we know $3 \leqslant a_{1} + b_{1} \leqslant 9$. Therefore, all pairs in $S$ that satisfy $(a+b) \mid a b$ are as follows:
$$
\begin{array}{l} 
a_{1}+b_{1}=3:(6,3),(12,6),(18,9),(24,12), \\
(30,15),(36,18),(42,21),(48,24) ; \\
a_{1}+b_{1}= 4:(12,4),(24,8),(36,12),(48,16) ; \\
a_{1}+b_{1}= 5:(20,5),(40,10),(15,10),(30,20), \\
(45,30) ; \\
a_{1}+b_{1}= 6:(30,6) ; \\
a_{1}+b_{1}= 7:(42,7),(35,14),(28,21) ; \\
a_{1}+b_{1}= 8:(40,24) ; \\
a_{1}+b_{1}= 9:(45,36) .
\end{array}
$$

There are 23 pairs in total.
Let
$$
M=\{6,12,15,18,20,21,24,35,40,42,45,48\},
$$

then $|M|=12$ and each of the 23 pairs 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
$$
\begin{array}{l}
(6,3),(12,4),(20,5),(42,7),(24,8),(18,9), \\
(40,10),(35,14),(30,15),(48,16),(28,21),(45,36)
\end{array}
$$

are disjoint and all satisfy the given condition. Therefore, for any 39-element subset of $S$, it is only 11 elements less than $S$, and these 11 elements can belong to at most 11 of the above 12 pairs, so there must be one of the 12 pairs that belongs to this 39-element subset.
In conclusion, the smallest natural number $k=39$.