*end Let $S=\{1,2, \cdots, 2005\}$. If any set of $n$ pairwise coprime numbers in $S$ contains at least one prime number, find the minimum value of $n$.

First, we have $n \geqslant 16$. In fact, take the set
$$
A_{0}=\left\{1,2^{2}, 3^{2}, 5^{2}, \cdots, 41^{2}, 43^{2}\right\},
$$

whose elements, except for 1, are all squares of prime numbers not exceeding 43. Then $A_{0} \subseteq S, \left|A_{0}\right|=15$, and any two numbers in $A_{0}$ are coprime, but there are no prime numbers in it, which shows that $n \geqslant 16$.

Second, we prove: For any $A \subseteq S, n=|A|=16$, if any two numbers in $A$ are coprime, then $A$ must contain a prime number.

Using proof by contradiction, assume that $A$ contains no prime numbers. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{16}\right\}$, and consider two cases. (16), so the prime factors of $a_{i}$ and $a_{j}$ are all different. Let the smallest prime factor of $a_{i}$ be $p_{i}$. Without loss of generality, assume $p_{1} < p_{2} < \cdots < p_{16}$. Then $a_{1} \geqslant p_{1}^{2} \geqslant 2^{2}, a_{2} \geqslant p_{2}^{2} \geqslant 3^{2}, \cdots, a_{16} \geqslant p_{16}^{2} \geqslant 53^{2}>2005$, a contradiction.
$$
(2) If $1 \in A$, then without loss of generality, let $a_{16}=1, a_{1}, \cdots, a_{15}$ be composite numbers. Similarly, let the smallest prime factor of $a_{i}$ be $p_{i}$. Then $a_{1} \geqslant p_{1}^{2} \geqslant 2^{2}, a_{2} \geqslant p_{2}^{2} \geqslant 3^{2}, \cdots, a_{15} \geqslant p_{15}^{2} \geqslant 47^{2}>2005$, a contradiction.
From (1) and (2), we know that the assumption is false, thus $A$ must contain a prime number, i.e., when $n=|A|=16$, the conclusion holds.

In summary, the minimum value of $n$ is 16.