Example 8 Let $S=\{1,2, \cdots, 98\}$. Find the smallest natural number $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which one number is coprime with the other 4 numbers, and in the other group, there is always one number that is not coprime with the other 4 numbers.
(1998, China Mathematical Olympiad)
Explanation: First, there are 49 even numbers in $S$, and the 49-element subset formed by them obviously does not meet the requirement. Therefore, the smallest natural number $n \geqslant 50$.

To prove that in any 50-element subset $T$ of $S$, there are 10 numbers that meet the requirement, we first prove a strengthened proposition:
In any 50-element subset $T$ of $S$, it is possible to select 10 numbers such that one of them is coprime with the other 9, and the other 9 numbers are pairwise not coprime.
$$
\begin{array}{l}
\text { Let } M_{1}=\{2 k \mid k=1,2, \cdots, 49\}, \\
M_{2}=\{6 k-31 k=1,2, \cdots, 16\}, \\
M_{3}=\{5,25,35,55,65,85,95,77,91\}, \\
M_{4}=\{7,49,11,13,17,19,23,29,31,37, \\
41,43,47\} . \\
M_{5}=\{1,53,59,61,67,71,73,79,83,89,97\} .
\end{array}
$$

Clearly, $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}$ are pairwise disjoint,
$$
\begin{aligned}
\bigcup_{i=1}^{5} M_{i}=S & \text { and } \\
\left|M_{1}\right| & =49,\left|M_{2}\right|=16,\left|M_{3}\right|=9, \\
\left|M_{4}\right| & =13,\left|M_{5}\right|=11 .
\end{aligned}
$$

Assume there exists $T \subset S$, such that $|T|=50$ and the conclusion of the strengthened proposition does not hold.
(1) If $\left|T \cap M_{2}\right| \geqslant 9$, by the contradiction assumption, the following 15 odd numbers
$$
\{37,41,43,47\} \cup M_{5}
$$

cannot be in $T$, so $T$ must contain at least 16 even numbers.
(2) When $\left|T \cap M_{2}\right| \leqslant 8$, at least 8 of the 16 odd numbers in $M_{2}$ are not in $T$, so $T$ must contain at least 9 even numbers. By the contradiction assumption, $T \cap M_{5}=\varnothing$, so $T$ must contain at least 20 even numbers.

Combining (1) and (2), it is clear that, in any case, $T$ contains at least 16 even numbers.
(3) When $\left|T \cap M_{1}\right| \geqslant 16$, because for each odd number in $M_{4} \cup$ $M_{5}$, the number of even numbers in $S$ that are not coprime with it does not exceed 7, these 24 odd numbers are not in $T$. Therefore, $T$ contains at least 25 even numbers.
(4) When $\left|T \cap M_{1}\right| \geqslant 25$, because for each odd number in $M_{3} \cup$ $M_{4} \cup M_{5}$, the number of even numbers that are not coprime with it does not exceed 15, by the contradiction assumption, they are not in $T$, so $T$ contains at least 34 even numbers.
(5) When $\left|T \cap M_{1}\right| \geqslant 34$, because for each odd number in $M_{2}$, the number of even numbers that are not coprime with it does not exceed 23, by the contradiction assumption, they are not in $T$, thus, $T$ contains no odd numbers, which is impossible.
In summary, the smallest natural number $n=50$.