Example 6 From the 100 positive integers $1,2,3, \cdots, 100$, if $n$ numbers are taken, among these $n$ numbers, there are always 4 numbers that are pairwise coprime. Find the minimum value of $n$.


Let $S=\{1,2,3, \cdots, n\}, A_{i}=\{k \mid k \in S$ and $k$ is divisible by $i\}(i=2,3,5)$, then by the principle of inclusion-exclusion, we have
$$
\begin{aligned}
\left|A_{2} \cup A_{3} \cup A_{5}\right| & =\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{5}\right|-\left|A_{2} \cap A_{3}\right|-\left|A_{2} \cap A_{5}\right|-\left|A_{3} \cap A_{5}\right|+\left|A_{2} \cap A_{3} \cap A_{5}\right| \\
& =\left[\frac{100}{2}\right]+\left[\frac{100}{3}\right]+\left[\frac{100}{5}\right]-\left[\frac{100}{2 \times 3}\right]-\left[\frac{100}{2 \times 5}\right]-\left[\frac{100}{3 \times 5}\right]+\left[\frac{100}{2 \times 3 \times 5}\right] \\
& =50+33+20-16-10-6+3=74 .
\end{aligned}
$$

In $A_{2} \cup A_{3} \cup A_{5}$, if we choose 4 numbers, at least 2 of them belong to the same set among $A_{2}, A_{3}, A_{5}$, and they are not coprime.
Therefore, the minimum value of $n$ is $\geqslant 75$.
Next, we prove that the minimum value of $n$ is 75.
Let $B_{1}=\{1\} \cup\{100$ prime numbers $\}, B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}\right\}$,
$$
B_{3}=\{2 \times 47,3 \times 31,5 \times 19,7 \times 13\}, B_{4}=\{2 \times 43,3 \times 29,5 \times 17,7 \times 11\},
$$

and let $B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4}$. Then $|B|=26+3 \times 4=38,|S \backslash B|=100-38=62$. Thus, when choosing 75 numbers from $S$, at least $75-62=13$ of them belong to $B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4}$. By the pigeonhole principle, among these 13 numbers, at least $\left[\frac{13-1}{4}\right]+1=4$ numbers belong to the same subset among $B_{1}, B_{2}, B_{3}, B_{4}$, and they are pairwise coprime.

In conclusion, the minimum value of $n$ is 75.