$2 \cdot 104$ Let $S=\{1,2, \cdots, 280\}$. Find the smallest natural number $n$, such that every $n$-element subset of $S$ contains 5 pairwise coprime numbers.
[Solution 1] Let
$$
A_{i}=\left\{i k \mid k=1,2, \cdots,\left[\frac{280}{i}\right]\right\}, i=1,2, \cdots \text {. }
$$

It is easy to calculate that $\left|A_{2}\right|=140,| A_{3}|=93,| A_{5}|=56,| A_{7}|=40,| A_{6} \mid=$ $\left|A_{2} \cap A_{3}\right|=46,\left|A_{10}\right|=\left|A_{2} \cap A_{5}\right|=28,\left|A_{14}\right|=\left|A_{2} \cap A_{7}\right|=$ $20,\left|A_{15}\right|=\left|A_{3} \cap A_{5}\right|=18,\left|A_{21}\right|=\left|A_{3} \cap A_{7}\right|=13,\left|A_{35}\right|=$ $\left|A_{5} \cap A_{7}\right|=8,\left|A_{30}\right|=\left|A_{2} \cap A_{3} \cap A_{5}\right|=9,\left|A_{42}\right|=\mid A_{2}$ $\cap A_{3} \cap A_{7}|=6,| A_{70}|=| A_{2} \cap A_{5} \cap A_{7}|=4,| A_{105}|=| A_{3} \cap A_{5} \cap$ $A_{7}|=2,| A_{210}|=| A_{2} \cap A_{3} \cap A_{5} \cap A_{7} \mid=1$. Let $A=A_{2} \cup A_{3} \cup$ $A_{5} \cup A_{7}$, then by the principle of inclusion and exclusion, we can calculate
$$
|A|=216 \text {. }
$$

Since when choosing 5 numbers from $A$, there must be two numbers in the same $A_{i}(i \in\{2,3,5,7\})$, which are not coprime, it follows that the smallest natural number $n \geqslant 217$.

On the other hand, let $T \subset S$ and $|T|=217$. Denote the set of all primes and 1 in $S$ as $M$, then $|M|=60$.
(1) If $|T \cap M| \geqslant 5$, then the problem is solved.
(2) If $|T \cap M|=4$, let the remaining primes be arranged in ascending order as $p_{1}, p_{2}$, $p_{3}, \cdots$, obviously $p_{1} \leqslant 11, p_{2} \leqslant 13, p_{3} \leqslant 17, p_{4} \leqslant 19, p_{5} \leqslant 23$. Then we have
$$
\left\{p_{1}^{2}, p_{1} p_{2}, p_{1} p_{3}, p_{1} p_{4}, p_{1} p_{5}, p_{2}^{2}, p_{2} p_{3}, p_{2} p_{4}\right\} \subset S .
$$

Since there are 220 composite numbers in $S$ and at this time $T$ contains 213 composite numbers, there are only 7 composite numbers in $S-T$, so among the 8 elements of the above set, there must be one contained in $T$, which together with the 4 numbers in $T \cap M$ meets the requirements of the problem.
(3) Suppose $|T \cap M| \leqslant 3$. At this time, at most 6 composite numbers in $S$ are not in $T$. If 5 or 4 elements of the set $\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}$ are contained in $T$, then the problem is reduced to the first two cases. Suppose at most 3 of these 6 composite numbers are contained in $T$, then at most 3 other composite numbers are not in $T$. Therefore, among the following two sets
$$
\begin{array}{l}
\{2 \times 41,3 \times 37,5 \times 31,7 \times 29,11 \times 23,13 \times 19\}, \\
\{2 \times 37,3 \times 31,5 \times 29,7 \times 23,11 \times 19,13 \times 17\}
\end{array}
$$

at most 3 of the 12 composite numbers are not in $T$. By the pigeonhole principle, there must be a set of at least 5 numbers contained in $T$. Clearly, these 5 numbers are pairwise coprime.
In summary, the smallest natural number $n=217$.
[Solution 2] From the first part of Solution 1, we know that the smallest natural number $n \geqslant 217$. Now we prove the latter part. Let
$B_{1}=\{1$ and all primes in $S \}$,
$B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}$,
$B_{3}=\{2 \times 41,3 \times 37,5 \times 31,7 \times 29,11 \times 23,13 \times 19\}$,
$B_{4}=\{2 \times 37,3 \times 31,5 \times 29,7 \times 23,11 \times 19,13 \times 17\}$,
$B_{5}=\{2 \times 31,3 \times 29,5 \times 23,7 \times 19,11 \times 17\}$,
$B_{6}=\{2 \times 29,3 \times 23,5 \times 19,7 \times 17,11 \times 13\}$,

and let $B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4} \cup B_{5} \cup B_{6}$. Since $\left|B_{1}\right|=60$ and $B_{i} \cap B_{j}=\varnothing, 1 \leqslant i<j \leqslant 6$, we have $|B|=88$. Thus, $|S-B|=192$.

For any $T \subset S,|T|=217$, we have $|T \cap B| \geqslant 25$, i.e., $T$ contains at least 25 elements of $B$. Therefore, by the pigeonhole principle, there exists $i_{0}, 1 \leqslant i_{0} \leqslant 6$, such that $B_{i_{0}}$ contains at least 5 elements of $T$. Clearly, these 5 numbers are pairwise coprime. Therefore, the smallest natural number $n=217$.
[Solution 3] From the first part of Solution 1, we know that the smallest natural number $n \geqslant 217$. Now we prove the latter part. Let the set of 1 and all primes in $S$ be $M$, then $|M|=60$. For any $T \subset S,|T|=217$, if $|T \cap M| \geqslant 5$, then the problem is naturally solved. Suppose $|T \cap M| \leqslant 4$, then at most 7 composite numbers in $S$ are not in $T$. Let
$$
\begin{array}{l}
M_{1}=\left\{2 \times 23,3 \times 19,5 \times 17,7 \times 13,11^{2}\right\}, \\
M_{2}=\{2 \times 29,3 \times 23,5 \times 19,7 \times 17,11 \times 13\}, \\
M_{3}=\{2 \times 31,3 \times 29,5 \times 23,7 \times 19,11 \times 17\}, \\
M_{4}=\{2 \times 37,3 \times 31,5 \times 29,7 \times 23,11 \times 19\}, \\
M_{5}=\{2 \times 41,3 \times 37,5 \times 31,7 \times 29,11 \times 23\}, \\
M_{6}=\{2 \times 43,3 \times 41,5 \times 37,7 \times 31,13 \times 17\}, \\
M_{7}=\{2 \times 47,3 \times 43,5 \times 41,7 \times 37,13 \times 19\}, \\
M_{8}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 13^{2}\right\} .
\end{array}
$$

Since at most 7 elements of these 8 sets are not in $T$, by the pigeonhole principle, there must be one set whose 5 elements are all in $T$. They are clearly pairwise coprime. Therefore, the smallest natural number $n=217$.