First we prove $n \ge 16$. In fact, let
$$A_0 = \{1, 2^2, 3^2, 5^2, \dots, 41^2, 43^2\},$$
where the members in $A_0$, other than $1$, are the squares of prime numbers not greater than $43$. Then $A_0 \subseteq S$, $|A_0| = 15$ and the numbers in $A_0$ are pairwise coprime but $A_0$ contains no prime number. Thus $n \ge 16$.

Next we show that for arbitrary $A \subseteq S$ with $n = |A| = 16$, if the numbers in $A$ are pairwise coprime, then $A$ must contain a prime number.
In fact, if $A$ contains no prime number, denote $A = \{a_1, a_2, \dots, a_{16}; a_1 < a_2 < \dots < a_{16}\}$. Then there are two possibilities.

(1) If $1 \notin A$, then $a_1, a_2, \dots, a_{16}$ are composite numbers. Since $(a_i, a_j) = 1$ ($1 \le i < j \le 16$), the prime factors of $a_i$ and $a_j$ are pairwise distinct. Let $p_i$ be the smallest prime factor of $a_i$. We may assume that $p_1 < p_2 < \dots < p_{16}$, then
$$a_1 \ge p_1^2 \ge 2^2,\quad a_2 \ge p_2^2 \ge 3^2,\quad \dots,\quad a_{15} \ge p_{15}^2 \ge 47^2 > 2005,$$
it leads to a contradiction.

(2) If $1 \in A$, let $a_{16} = 1$, $a_1, a_2, \dots, a_{15}$ are composite numbers. By the same assumption and argument of (1), we have $a_1 \ge p_1^2 \ge 2^2$, $a_2 \ge p_2^2 \ge 3^2$, $\dots, a_{15} \ge p_{15}^2 \ge 47^2 > 2005$. Again, it leads to a contradiction.

From (1) and (2), $A$ contains at least one prime number, i.e. when $n = |A| = 16$, the conclusion is true. Thus, the minimum number of $n$ is $16$.