(2) 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 $(a+b) \mid a b$.
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 $(a_{1}, b_{1})=1$, hence 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
$$
\left(a_{1}+b_{1}\right) \mid c
$$

Since $a, b \in S$, we have $a+b \leqslant 99$, i.e., $c\left(a_{1}+b_{1}\right) \leqslant 99$. Therefore, $3 \leqslant a_{1}+b_{1} \leqslant 9$. From this, we know that there are 23 pairs of distinct numbers in $S$ that satisfy the condition $(a+b) \mid a b$, as follows:
$$
a_{1}+b_{1}=3 ;(6,3),(12,6),(18,9),(24,12),(30,15),(36,18), (42,21),(48,24)
$$
$$
\begin{array}{l}
a_{1}+b_{1}=4:(12,4),(24,8),(36,12),(48,16) \\
a_{1}+b_{1}=5:(20,5),(40,10),(15,10),(30,20),(45,30) \\
a_{1}+b_{1}=6:(30,6) \\
a_{1}+b_{1}=7:(42,7),(35,14),(28,21) \\
a_{1}+b_{1}=8:(40,24) \\
a_{1}+b_{1}=9:(45,36)
\end{array}
$$
$$
\text{Let } M=\{6,12,15,18,20,21,24,35,40,42,45,48\} \text{, then }|M|=
$$
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. Thus, 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 missing only 11 elements from $S$, and these 11 elements can belong to at most 11 of the 12 pairs above, so there must be one of the 12 pairs that belongs to $R$.
In summary, the smallest natural number $k=39$.
Solution 1 When $n=3,4$, the conclusion is obviously true.
When $n=5$, in a regular pentagon with 5 diagonals, there are exactly 5 obtuse triangles and 5 acute triangles. When 5 magpies $A, B, C, D, E$ fly and land in the states shown in Figure 1, the acute triangles and obtuse triangles transform into each other, not satisfying the given condition.