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 $n > m$.