$2 \cdot 122$ 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 evenly divided into two groups, there is always a number in one group that is coprime with the other 4 numbers, and a number in the other group that is not coprime with the other 4 numbers.
[Solution] First, there are 49 even numbers in $S$, and the 49-element subset formed by them cannot meet the requirements of the problem, so the smallest natural number $n \geqslant 50$.

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

Obviously, $\left|M_{1}\right|=49,\left|M_{2}\right|=16,\left|M_{3}\right|=9,\left|M_{4}\right|=13,\left|M_{5}\right|=11, M_{i}$ $\cap M_{j}=\emptyset, i \neq j$ and $M_{1} \cup M_{2} \cup M_{3} \cup M_{4} \cup M_{5}=S$.
Assume there exists $T \subset S,|T|=50$ and the conclusion of the strengthened proposition does not hold.
(1) If $\left|T \cap M_{2}\right| \geqslant 9$, then by the contradiction assumption, the following 15 odd numbers
$$
\{37,41,43,47\} \cup M_{5}
$$

are not in $T$, so $T$ contains at least 16 even numbers.
(2) If $\left|T \cap M_{2}\right| \leqslant 8$, then at least 8 odd numbers in $M_{2}$ are not in $T$. Thus, $T$ contains at least 9 even numbers. By the contradiction assumption, $T \cap M_{5}=\emptyset$, so $T$ contains at least 20 even numbers.
(3) If $\left|T \cap M_{1}\right| \geqslant 16$, 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, so these 24 odd numbers are not in $T$, thus $T$ contains at least 25 even numbers.
(4) If $\left|T \cap M_{1}\right| \geqslant 25$. Because for each odd number in $M_{3} \cup M_{4} \cup M_{5}$, the number of even numbers in $S$ that are not coprime with it does not exceed 15, so these 33 odd numbers are not in $T$. Thus, $T$ contains at least 34 even numbers.
(5) If $\left|T \cap M_{1}\right| \geqslant 34$. Because for each odd number in $M_{2}$, the number of even numbers in $S$ that are not coprime with it does not exceed 23, thus each odd number in $S$ is the same. Hence, by the contradiction assumption, they are not in $T$. This leads to $|T| \leqslant 49$, a contradiction.
In summary, the smallest natural number $n=50$.