Team selection test for 50. IMO

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.