Solution: For each rational $\frac{a}{b}$ with $(a,b)=1$ prime $p$:$p|\frac{a}{b}$ <=>$p|a$(which also mean $(p,b)=1$) Assume that this sequence have a rational element,that is $\frac{s}{t}$ with $(s,t)=1$ and $t>1$,let $p$ be a prime divisor of $n$ First note that product of $n$ numbers of $M$ is an integer So let $c_1,c_2,...,c_{n.(n-1)}$.Consider pair $( \frac{s}{t};c_{(n-1)k+1};...;c_{(n-1)k+(n-1)})$ for all $k=0,n-1$ Then for each k,exist $1=<t_k=<n-1$ $c_{(n-1)k+t_k}$ is divisible by p Consider $n$ numbers,$c_{(n-1)k+t_k}$ for $k=0,n-1$,product of them is integer divisible by $p^n$,contradiction

Let $\frac{a}{b}\in M$. Choose a prime $p\mid b$. Note that at most $n-2$ elements $m\in M$ not including $\frac{a}{b}$ satisfy $v_p(m) \le 0$, because if $n-1$ elements existed we could choose those elements as well as $\frac{a}{b}$ to get a product that is not an integer. Thus, an infinite number of elements $m\in M$ satisfy $v_p(m) > 0$. Choose any $n$ of these elements to get a product with $v_p\left(\prod m\right) \ge n$, contradiction. Thus, our initial assumption is incorrect, so $M$ contains only integers.