, 2008

Solution:
We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\frac{p_{1}}{q_{1}}, \frac{p_{2}}{q_{2}}, \ldots, \frac{p_{k}}{q_{k}}, \ldots$, with $(p_{k}, q_{k})=1$ and $q_{k}>1$ for each $k$. Let $\frac{p}{q}=\frac{p_{1} p_{2} \ldots p_{n-1}}{q_{1} q_{2} \ldots q_{n-1}}$, where $(p, q)=1$. For each $i \geq n$, the number $\frac{p}{q} \cdot \frac{p_{i}}{q_{i}}$ is an integer, so $q_{i}$ is a divisor of $p$ (as $q_{i}$ and $p_{i}$ are coprime). But $p$ has a finite set of divisors, so there are $n$ numbers of $M$ with equal denominators. Their product cannot be an integer, a contradiction.

Now suppose that $M$ contains a fraction $\frac{a}{b}$ in lowest terms with $b>1$. Take a prime divisor $p$ of $b$. If we take any $n-1$ integers from $M$, their product with $\frac{a}{b}$ is an integer, so some of them is a multiple of $p$. Therefore there are infinitely many multiples of $p$ in $M$, and the product of $n$ of them is not $n$-free, a contradiction.