Four, (15 points) Given a positive integer $n$ that satisfies the following condition: among any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$.


Because $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor no greater than 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17,19,23$,
$29,31,37,41,43$.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive integers greater than 1 and not exceeding 2009 that are pairwise coprime. If they are all composite, let $p_{i}$ be the prime factor of $a_{i}$ that does not exceed 44.

Since $a_{1}, a_{2}, \cdots, a_{n}$ are pairwise coprime, $p_{1}, p_{2}, \cdots, p_{n}$ are all distinct.
Thus, $n \leqslant 14$.
In other words, when $n \geqslant 15$, at least one of $a_{1}, a_{2}, \cdots, a_{n}$ must be a prime number.

On the other hand, $2^{2}, 3^{2}, 5^{2}, \cdots, 43^{2}$ are 14 positive integers greater than 1 and not exceeding 2009 that are pairwise coprime and contain no prime numbers.
Therefore, the minimum value of $n$ is 15.