N35 (32-3, China) Let $S=\{1,2,3, \cdots, 280\}$, find the smallest positive integer $n$, such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers.
Solve First estimate the lower bound of $n$.
Let $A_{i}(i=2,3,5,7)$ be the set of positive integers in $S$ that are divisible by $i$, and $A=A_{2} \cup A_{3} \cup A_{5} \cup A_{7}$.
By the principle of inclusion-exclusion, we have
$$
\begin{aligned}
|A|= & \left|A_{2}\right|+\left|A_{3}\right|+\left|A_{5}\right|+\left|A_{7}\right|-\left|A_{2} \cap A_{3}\right|-\left|A_{2} \cap A_{5}\right| \\
& -\left|A_{2} \cap A_{7}\right|-\left|A_{3} \cap A_{5}\right|-\left|A_{3} \cap A_{7}\right|-\left|A_{5} \cap A_{7}\right| \\
& +\left|A_{2} \cap A_{3} \cap A_{5}\right|+\left|A_{2} \cap A_{3} \cap A_{7}\right|+\left|A_{2} \cap A_{5} \cap A_{7}\right| \\
& +\left|A_{3} \cap A_{5} \cap A_{7}\right|-\left|A_{2} \cap A_{3} \cap A_{5} \cap A_{7}\right| \\
= & {\left[\frac{280}{2}\right]+\left[\frac{280}{3}\right]+\left[\frac{280}{5}\right]+\left[\frac{280}{7}\right]-\left[\frac{280}{2 \times 3}\right]-\left[\frac{280}{2 \times 5}\right] } \\
& -\left[\frac{280}{2 \times 7}\right]-\left[\frac{280}{3 \times 5}\right]-\left[\frac{280}{3 \times 7}\right]-\left[\frac{280}{5 \times 7}\right]+\left[\frac{280}{2 \times 3 \times 5}\right] \\
& +\left[\frac{280}{2 \times 3 \times 7}\right]+\left[\frac{280}{2 \times 5 \times 7}\right]+\left[\frac{280}{3 \times 5 \times 7}\right]-\left[\frac{280}{2 \times 3 \times 5 \times 7}\right] \\
= & 140+93+56+40-46-28-20-18-13-8+9+6+4 \\
& +2-1 \\
= & 216,
\end{aligned}
$$

In $S$, take the set $A$ with 216 elements. By the pigeonhole principle, any 5 of these 216 elements must have two that belong to one of the sets $A_{2}, A_{3}, A_{5}, A_{7}$, so they are not coprime. That is, from these 216 numbers, it is impossible to find 5 numbers that are pairwise coprime. Therefore, $n \geqslant 217$.
Next, we will prove that in any 217 elements of $S$, there must be 5 numbers that are pairwise coprime. Let
$$
\begin{array}{l}
B_{1}=\{1 \text { and all primes in } S\}, \\
B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}, \\
B_{3}=\{2 \times 131,3 \times 89,5 \times 53,7 \times 37,11 \times 23,13 \times 19\}, \\
B_{1}=\{2 \times 127,3 \times 83,5 \times 47,7 \times 31,11 \times 19,13 \times 17\}, \\
B_{5}=\{2 \times 113,3 \times 79,5 \times 43,7 \times 29,11 \times 17\}, \\
B_{6}=\{2 \times 109,3 \times 73,5 \times 41,7 \times 23,11 \times 13\}, \\
B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4} \cup B_{5} \cup B_{6} .
\end{array}
$$

It is easy to see that the numbers in $B_{i}(i=1,2, \cdots, 6)$ are pairwise coprime, and $B_{i} \cap B_{j}=\varnothing(1 \leqslant i<j \leqslant 6)$. $\left|B_{1}\right|=60,\left|B_{2}\right|=\left|B_{3}\right|=\left|B_{4}\right|=6,\left|B_{3}\right|=\left|B_{6}\right|=5$, so
$$
\begin{array}{c}
|B|=60+6 \times 3+5 \times 2=88 . \\
|S \backslash B|=280-88=192 .
\end{array}
$$
In any 217 elements of $S$, there are at least $217-192=25$ elements in $B$. By the pigeonhole principle, among these 25 elements, there are at least $\left[\frac{25}{6}\right]+1=5$ elements in some $B_{i}(1 \leqslant i \leqslant 6)$, and these 5 elements are pairwise coprime.
In summary: The minimum value of $n$ is 217.