6. 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 35 prime numbers).
6. Considering the number of multiples of 2, 3, or 5 in $S$, we have:
$$
\left[\frac{150}{2}\right]+\left[\frac{150}{3}\right]+\left[\frac{150}{5}\right]+\left[\frac{150}{2 \cdot 3}\right]-\left[\frac{150}{2 \cdot 5}\right]-\left[\frac{150}{3 \cdot 5}\right]+
$$
$$
\left[\frac{150}{2 \cdot 3 \cdot 5}\right]=110 \text {. }
$$

When $n=110$, we can take all the multiples of 2, 3, or 5, so in this subset, it is impossible to find 4 pairwise coprime numbers. Therefore, $n \geqslant 111$. We now prove that $n=111$ is feasible.

To do this, we construct the following 6 sets: $A_{1}=\{1\} \cup\{S$ with 35 prime numbers $\}, A_{2}=\{2 \cdot 67,3 \cdot 43,5 \cdot 17,7 \cdot 19,11 \cdot 13 \mid$, $A_{3}=\{2 \cdot 53,3 \cdot 47,5 \cdot 23,7 \cdot 17\}, A_{4}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}\right\}, A_{5}=\left\{2 \cdot 19,3^{3}, 5 \cdot 13,7 \cdot 11\right\}, A_{6}=\left\{2^{3}\right.$, $3 \cdot 23,5 \cdot 11,7 \cdot 13 \mid$. Let $A=A_{1} \cup A_{2} \cup \cdots \cup A_{6}$. Each of the above $A_{i}(i=1,2, \cdots, 6)$ must contain at least 4 elements that are pairwise coprime, and $A_{i} \cap A_{j}=\varnothing(1 \leqslant i<j \leqslant 60)$. To prove the conclusion of this problem, we must ensure that $|A|=58$. Thus, if 111 numbers are selected from $S$, at least 19 numbers must be selected from $A$. 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.