Let $a_1, \dots, a_{2008} \in M$ and $A = a_1 \dots a_{2008} = \frac{p}{q}$, $(p, q) = 1$. Assume that $M$ contains infinitely many numbers $\alpha_i = \frac{p_i}{q_i}$ such that $(p_i, q_i) = 1$, $q_i > 1$ and $\alpha_i \ne a_1, \dots, a_{2008}$. Since $\alpha_i p$ is an integer, then $q_i$ divides $p$ and hence infinitely many of the numbers $q_i$ are equal. Then the product of 2009 of the respective $\alpha_j$ is not an integer, a contradiction. In particular, $M$ contains infinitely many integers.

Assume now that $\frac{a}{b} \in M$, $(a, b) = 1$ and $b > 1$. If $p$ is a prime divisor of $b$, then the given condition easily implies that $p$ divides infinitely many integers in $M$. Then the product of any 2009 of them is divisible by $p^{2009}$, a contradiction.