49th Mathematical Olympiad in Ukraine

Let $M_1=\{a,b,c,d\}$ be a four-element subset of $M$ such that $abcd<2009$. Then the subset $M_2=\{\frac{48}{a},\frac{48}{b},\frac{48}{c},\frac{48}{d}\}$ is also a four-element subset of $M$ because all of its elements are distinct and belong to $M$ and $\frac{48}{a}\cdot \frac{48}{b}\cdot \frac{48}{c}\cdot \frac{48}{d}=\frac{48^4}{abcd}>\frac{48^4}{2009}>2009$. It follows that $n\ge m$. But also we have the subset $\{1,2,24,48\}$ which has product of elements $1\cdot 2\cdot 24\cdot 48=48^2>2009$ and doesn't have any corresponding pair with product of elements less than $2009$. Thus $n>m$.