Example 8- 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.
Thought analysis: First, construct a subset of $S$ to obtain a lower bound for $n$; then, by constructing another type of subset of $S$, show that the above lower bound for $n$ can be achieved, thereby determining the minimum value of $n$.

Solution: Let $A_{1}=\{S$ of numbers divisible by 2 $\}, A_{2}=\{S$ of numbers divisible by 3 $\}, A_{3}=\{S$ of numbers divisible by 5 $\}, A_{4}=\{S$ of numbers divisible by 7 $\}$, and denote $A=A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$. Using the principle of inclusion-exclusion, we calculate $|A|=216$.

Since any 5 numbers chosen from $A$ must have at least two numbers in the same $A_{i}$ $(i=1,2,3,4)$, and these two numbers are not coprime, it follows that $n \geqslant 217$.

On the other hand, let $B_{1}=\{1$ and all prime numbers in $S\}$, $B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}$,
$$
B_{3}=\{2 \times 131,3 \times 89,5 \times 53,7 \times 37,11 \times
$$
$23,13 \times 19\}$
$$
B_{4}=\{2 \times 127,3 \times 83,5 \times 47,7 \times 31,11 \times
$$
$19,13 \times 17\}$
$$
B_{5}=\{2 \times 113,3 \times 79,5 \times 43,7 \times 29,11 \times
$$
17\}
$$
B_{6}=\{2 \times 109,3 \times 73,5 \times 41,7 \times 23,11 \times
$$
$13\}$,
denote $B=B_{1} \cup B_{2} \cup \cdots \cup B_{6}$, then $|B|=88$, so $S-B$ contains 192 elements. When choosing 217 numbers from $S$, since $217-192=25$, at least 25 elements must be in $B$, and $25=4 \times 6+1$, so among these 25 elements, there must be 5 numbers belonging to the same $B_{i}$, and these numbers are clearly pairwise coprime.
Therefore, the minimum value of $n$ is 217.