Example 6 Let $S=\{1,2,3, \cdots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers.
Let $A_{1}=\{S$ where the numbers are divisible by 2 $\}, A_{2}=\{S$ where the numbers are divisible by 3 $\}, A_{3}=\{S$ where the numbers are divisible by 5 $\}, A_{4}=\{S$ where the numbers are divisible by 7 $\}$, and denote $A=A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$. It is easy to see that $|A|=216$.
Since any 5 numbers chosen from $A$ must have at least two numbers in the same $A_{i}$, 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 $S\}, B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}, B_{3}=\{2 \times$
$$
\begin{array}{l}
131,3 \times 89,5 \times 53,7 \times 37,11 \times 23,13 \times 19\}, B_{4}=\{2 \times 127,3 \times 83,5 \times 47,7 \times 31,11 \times 19, \\
13 \times 17\}, B_{5}=\{2 \times 113,3 \times 79,5 \times 43,7 \times 29,11 \times 17\}, B_{6}=\{2 \times 109,3 \times 73,5 \times 41,7 \times \\
23,11 \times 13\} .
\end{array}
$$

Denote $B=B_{1} \cup B_{2} \cup \cdots \cup B_{6}$, then $|B|=88$. Thus, $S / B$ contains 192 elements. If we choose 217 numbers from $S$, since $217-192=25$, there must be 25 elements in $B$. Therefore, there must be 5 numbers in the same $B_{i}$, and these numbers are pairwise coprime.
Hence, the minimum value of $n$ is 217.