Find the smallest positive integer $n$, such that every $n$-element subset of $S=\{1,2, \cdots, 150\}$ contains 4 pairwise coprime numbers (it is known that $S$ contains a total of 35 prime numbers).
Solution: Consider the number of multiples of 2, 3, or 5 in $S$. We have
$$
\begin{array}{l}
{\left[\frac{150}{2}\right]+\left[\frac{150}{3}\right]+\left[\frac{150}{5}\right]-\left[\frac{150}{2 \times 3}\right]-\left[\frac{150}{2 \times 5}\right]} \\
-\left[\frac{150}{3 \times 5}\right] +\left[\frac{150}{2 \times 3 \times 5}\right]=110 .
\end{array}
$$

When $n=110$, all numbers can be multiples of 2, 3, or 5, so in this subset, it is impossible to find 4 numbers that are pairwise coprime. Therefore, $n \geqslant 111$.
Next, we prove that $n=111$ satisfies the requirement.
To do this, construct the following 6 sets:
$$
\begin{array}{l}
A_{1}=\{1\} \cup\{35 \text{ primes in } S\}, \\
A_{2}=\{2 \times 67,3 \times 43,5 \times 17,7 \times 19,11 \times 13\}, \\
A_{3}=\{2 \times 53,3 \times 47,5 \times 23,7 \times 17\} . \\
A_{4}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}\right\}, \\
A_{5}=\left\{2 \times 19,3^{3}, 5 \times 13,7 \times 11\right\}, \\
A_{6}=\left\{2^{3}, 3 \times 23,5 \times 11,7 \times 13\right\} . \\
\text{Let } A=A_{1} \cup A_{2} \cup A_{3} \cup \cdots \cup A_{6} .
\end{array}
$$

Each of the above $A_{i}(i=1,2, \cdots, 6)$ must contain at least 4 elements that are pairwise coprime, and $A_{i} \cap A_{j}=\varnothing(1 \leqslant i<j \leqslant 6)$, to prove the conclusion of this problem, it is necessary to ensure that $|A|=58$.

Thus, if 111 numbers are selected from $S$, at least 19 numbers must be selected from $A$, and by the pigeonhole principle, there must be some $A_{i}(i=1,2, \cdots, 6)$ from which 4 numbers are selected, and these four numbers are pairwise coprime.
(Yuan Jin, Anhui Normal University Affiliated High School, 241001)