29.3.14 Xuan Yuan seeks the smallest natural number $n$ with the property that any $n$-element subset of $\{1,2, \cdots, 60\}$ must contain three numbers that are pairwise coprime.
Parse and extract all multiples of 2 and 3 from 1 to 60, a total of $\left[\frac{60}{2}\right]+\left[\frac{60}{3}\right]-\left[\frac{60}{2 \times 3}\right]=40$ numbers, none of these 40 numbers are pairwise coprime.
Let $A_{1}=\{1$ and all prime numbers $\}, A_{2}=\left\{2 \times 29,3 \times 19,5 \times 11,7^{2}\right\}$,
$A_{3}=\{2 \times 23,3 \times 17,5 \times 7\}, A_{4}=\left\{2 \times 19,3 \times 11,5^{2}\right\}, A_{5}=\{$ other numbers $\}$, then $\left|A_{1} \cup A_{2} \cup A_{3} \cup A_{4}\right|=28$, if 41 numbers are taken, at least $41-$ $(60-28)=9$ numbers are taken from these 28 numbers, by the pigeonhole principle, at least one $A_{i}(i=1,2,3,4)$ will have 3 numbers taken, and these 3 numbers are pairwise coprime.
In summary, the minimum value of $n$ is 41.