11. (China Mathematical Olympiad) Let $S=\{1,2,3, \cdots, 98\}$, 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 a number in one group that is coprime with the other four numbers, and a number in the other group that is not coprime with the other four numbers.
11. Proof: Take the set $E=\{2 x \mid x \in \mathbf{N}$, and $1 \leqslant x \leqslant 49\}$ of 49 even numbers from $S$. Then, for any $k(k \geqslant 10)$ numbers taken from $E$, it is impossible to select 10 numbers that meet the requirements. Therefore, $n \geqslant 50$.
For any odd number $x$ in $S$, the number of even numbers in $S$ that are not coprime with $x$ is $f(x) \leqslant\left[\frac{49}{3}\right]=16$.
Let $T$ be any 50-element subset of $S$, with $\alpha$ even numbers and $50-\alpha=\beta$ odd numbers. Since $|E|=49$, $T$ must contain at least one odd number.
The set of numbers in $S$ whose smallest prime factor is $k$ (where $k$ is a prime number) is denoted as $R_{k}$, then
$$
\begin{array}{l}
R_{3}=\{3 \times 1,3 \times 3, \cdots, 3 \times 31\},\left|P_{3}\right|=16 ; \\
R_{5}=\{5 \times 1,5 \times 5,5 \times 7,5 \times 11,5 \times 13,5 \times 15,5 \times 19\},\left|P_{5}\right|=7, \\
R_{7}=\{7 \times 1,7 \times 7,7 \times 11,7 \times 13\},\left|P_{7}\right|=4,
\end{array}
$$
$\left|R_{k}\right|=1$ (where $k$ is a prime number greater than 10).
If there exists an odd number $x$ in $T$ such that the number of even numbers in $T$ that are coprime with $x$ is $\geqslant 9$, then $T$ meets the requirements of the problem.
(1) If $\alpha \geqslant 25$,

then since $f(x) \leqslant 16$, for any odd number $a$ in $T$, the number of even numbers in $T$ that are coprime with $a$ is $\geqslant 25-16=9$. The problem is proved.
(2) If $16 \leqslant \alpha \leqslant 24$, then $\beta \geqslant 26$.

Since $\left|\{1\} \cup R_{3} \cup R_{5}\right|=1+16+7=24<\beta$, there must be at least one odd number $a \in R_{7} \cup R_{k}$ in $T$, and $f(a) \leqslant f(7)=\left[\frac{49}{7}\right]=7$. Thus, the number of even numbers in $T$ that are coprime with $a$ is $\geqslant 16-7=9$. The problem is proved.
(3) If $10 \leqslant \alpha \leqslant 15$, then $\beta \geqslant 35$.
Since $\left|\{1\} \cup R_{3} \cup R_{5} \cup R_{7} \cup\{11,13,17,19,23,29\}\right|=34<\beta$, there must be at least one odd number $a \geqslant 31$ in $T$, and $f(a)=1$. Thus, the number of even numbers in $T$ that are coprime with $a$ is $\geqslant 10-1=9$. The problem is proved.
(4) If $\alpha=9$, then $\beta=41$,
Since $\left|\{1\} \cup R_{3} \cup R_{5} \cup R_{7} \cup\{11,13,17,19,23,29,31,37,41,43,47\}\right|=39<\beta$, there must be at least one prime number $a \geqslant 53$ in $T$, and all 9 even numbers in $T$ are coprime with $a$. The problem is proved.
(5) If $\alpha \leqslant 8, \beta \geqslant 42$,
Since $\left|\{1\} \cup R_{3} \cup R_{5} \cup R_{7} \cup\{11,13,17,19,23,29,31,37,41,43,47\}\right|=39<\beta$, there must be at least one prime number $a \geqslant 53$ in $T$, and $a$ is coprime with all other numbers in $T$. The number of odd numbers in $S$ not in $T$ is $50-\beta \leqslant 7$. Taking all the odd numbers of the form $6k-3$ in $T$, there are at least $16-7=9$ such numbers, and they are not coprime. The problem is proved.
In summary, the proposition is true.