Example 3 Let the set $S=\{1,2, \cdots, 280\}$. Find the smallest positive integer $n$, such that every subset of $S$ with $n$ elements contains five pairwise coprime numbers.
(32nd IMO)
【Analysis】The required $n$ satisfies: there exists an $(n-1)$-element subset $A \subset S$, such that in any 5-element subset of $A$, there are two elements that are not coprime. $n-1$ is the maximum value for which there exists a subset $A$ satisfying the above condition.

If $A$ can be written as the union of four sets $A_{1}, A_{2}, A_{3}, A_{4}$, and each set contains elements that are pairwise not coprime, then by the pigeonhole principle, we know that in any 5-element subset of $A$, there are two elements in the same $A_{i}$, which are not coprime. A simple construction method for a set of elements that are pairwise not coprime is to take the set of integer multiples of a fixed number. Thus, let $A_{1}, A_{2}, A_{3}, A_{4}$ be the sets of positive integers with the smallest prime factors being $2, 3, 5, 7$ respectively, and not exceeding 280.
Applying the principle of inclusion-exclusion, we calculate:
$$
\left|A_{1} \cup A_{2} \cup A_{3} \cup A_{4}\right|=216.
$$

Therefore, we conjecture that $n=217$.
To prove that any 217-element subset $C$ of $S$ contains five pairwise coprime elements. This is a type of problem where the pigeonhole principle applies, requiring the construction of pigeonholes, with each pigeonhole containing numbers that are pairwise coprime:
$$
\begin{aligned}
B_{1}= & \{1\} \cup\{\text{ primes not exceeding } 280 \}, \\
B_{2}= & \{\text{ squares of primes not less than } 17 \}, \\
B_{3}= & \{2 \times 41,3 \times 37,5 \times 31,7 \times 29, \\
& 11 \times 23,13 \times 19\}, \\
B_{4}= & \{2 \times 37,3 \times 31,5 \times 29,7 \times 23, \\
& 11 \times 19,13 \times 17\}, \\
B_{5}= & \{2 \times 31,3 \times 29,5 \times 23,7 \times 19,11 \times 17\}, \\
B_{6}= & \{2 \times 29,3 \times 23,5 \times 19,7 \times 17,11 \times 13\}.
\end{aligned}
$$

Thus, $B_{1} \cup B_{2} \cup \cdots \cup B_{6}$ includes all numbers not in $A = A_{1} \cup A_{2} \cup \cdots \cup A_{4}$, and each $B_{i}$ contains exactly four elements from $A$. If there is a set $C$ containing at least 217 elements, $C$ contains more elements in some $B_{i}$ than $A$ does in $B_{i}$, i.e., $\left|C \cap B_{i}\right| \geqslant 5$, then $C$ contains five elements that are pairwise coprime.
Therefore, the smallest $n$ satisfying the problem's conditions is 217.
【Comment】In Example 3, the two sets of pigeonholes $A_{i} (i=1,2,3, 4)$ and $B_{j} (j=1,2, \cdots, 6)$ satisfy:
$$
\left|A_{i} \cap B_{j}\right|=1,\left|A_{i} \cap A_{i^{\prime}}\right|=0,\left|B_{j} \cap B_{j^{\prime}}\right|=0;
$$

The union of all $A_{i}$ and $B_{j}$ is the entire set $S$;
Elements in $A_{i}$ are pairwise satisfying a certain condition, while elements in $B_{j}$ are pairwise not satisfying this condition.

Regardless of how this condition is described, if such types of pigeonhole sets can be constructed, then the solution to the corresponding problem is the number of elements in the union of $A_{i}$ plus 1.

In Example 4, when using the pigeonhole principle, it is necessary to reasonably adjust the pigeonholes.