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 there is 1 number that is coprime with the other 4 numbers, and in the other group, there is always 1 number that is not coprime with the other 4 numbers.
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,
$$
\bigcup_{i=1}^{5} M_{i}=S \text { and }
$$
$$
\begin{array}{l}
\left|M_{1}\right|=49,\left|M_{2}\right|=16,\left|M_{3}\right|=9, \\
\left|M_{4}\right|=13,\left|M_{5}\right|=11 .
\end{array}
$$

Assume there exists $T \subset S$, such that $|T|=50$ and the conclusion of the strengthened proposition does not hold.
(1) If $|T \cap M_{2}| \geqslant 9$, 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 $|T \cap M_{2}| \leqslant 8$, 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 $|T \cap M_{1}| \geqslant 16$, 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 $|T \cap M_{1}| \geqslant 25$, 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 $|T \cap M_{1}| \geqslant 34$, 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 $n=50$.