Example 5 Given a positive integer $n$ that satisfies the following condition: In 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 $n$. ${ }^{[2]}$
Since $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor not exceeding 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17$,
$19,23,29,31,37,41,43$.
On one hand,
$2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}$,
$19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$

are 14 positive integers greater than 1 and not exceeding 2009 that are pairwise coprime, and none of them are prime.
Therefore, $n \geqslant 15$.
On the other hand, let $a_{1}, a_{2}, \cdots, a_{15}$ be 15 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 $a_{i}$ not exceeding 44 $(i=1,2, \cdots, 15)$.

By the pigeonhole principle, there exist $i, j (1 \leqslant i<j \leqslant 15)$ such that $p_{i}=p_{j}$.
Thus, $a_{i}$ and $a_{j}$ are not coprime, which is a contradiction.
Therefore, among $a_{1}, a_{2}, \cdots, a_{15}$, there exists a prime number.
In conclusion, the minimum value of $n$ is 15.