Let $S=\{1,2, \cdots, n\}$. Find the smallest natural number $n$, such that when $S$ is arbitrarily divided into two subsets, there is always one subset that contains two different numbers $a$ and $b$, satisfying $(a+b) \mid a b$.
Solution: Two distinct numbers $a$ and $b$ that satisfy $(a+b) \mid a b$ are called a "good pair." Clearly, the answer to this problem is only related to these good pairs and has nothing to do with numbers that do not appear in any good pair. Therefore, we can list all the good pairs within a certain range, for example, all good pairs within 50:
$$
\begin{array}{l}
(3,6),(6,12),(9,18),(12,24),(15,30), \\
(18,36),(21,42),(24,48) ; \\
(4,12),(8,24),(12,36),(16,48) ; \\
(5,20),(10,40),(10,15),(20,30),(30,45) ; \\
(6,30) ; \\
(7,42),(14,35),(21,28) ; \\
(24,40) ; \\
(36,45) .
\end{array}
$$

Starting from 1, we sequentially divide the natural numbers $1,2, \cdots$ into two groups $A$ and $B$, ensuring that no good pair appears in either group. We need to see up to which number this can be done:
$$
\begin{aligned}
A: & 1,2,3,12,30,10,18,5,8,13,14,19,21,22, \\
& 23,31,32,33,34, \\
B: & 6,4,24,36,15,20,9,7,11,16,17,35,28, \\
& 25,26,27,29,37,38,39 .
\end{aligned}
$$

The above division already separates $\{1,2, \cdots, 39 \}$ into two groups with no good pairs in either group. However, it is impossible to add 40 to either group, as both $(10,40)$ and $(24,40)$ are good pairs. Therefore, the smallest natural number $n \geqslant 40$.

When $n=40$, consider the circle formed by the 7 numbers in Figure 4, where each adjacent pair of numbers is a good pair. These 7 numbers are divided into two sets, and there must be adjacent numbers in the same set, which of course form a good pair in that set.

In summary, the smallest natural number $n=40$.