Mongolian Mathematical Olympiad

Proceeding by contradiction, suppose that there exists $M$ such that $(n;[a_{1}n]+[a_{2}n]+\cdots +[a_{2014n}])\ne 1$ for all $n\ge M$. Consequently, for any prime $p_n$, (for all $p_n\ge M$) there exists $x_n\in \mathbb{N}$ such that: $[a_1{p}_n]+[a_2{p}_n]+\cdots +[a_{2014p_n}]=p_n{x}_n$. It is obvious that when $n\to \infty$ the sequence $x_n=\frac{[a_1{p}_n]+[a_2{p}_n]+\cdots +[a_{2014p_n}]}{p_n}$ converges to $a_1+a_2+\cdots +a_{2014}$ and because all terms of the sequence $(x_n{)}_{n\ge N}$ are integers, there exists $P$ such that $\forall n\ge P:x_n=a_1+a_2+\cdots +a_{2014}$. Since the latter is equivalent to $\forall n\ge P:\{a_1{p}_n\}+\{a_2{p}_n\}+\cdots +\{a_{2014p_n}\}=0$, we conclude that $\forall n\ge P:a_i{p}_n,i=1,2,\dots ,2014$ are integers. It leads to that $a_i,i=1,2,\dots ,2014$ are integers and but this is a contradiction.