Mongolian Mathematical Olympiad

Proceeding by contradiction, suppose that there exists $M$ such that $(n; [a_1n] + [a_2n] + \dots + [a_{2014n}]) \neq 1$ 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: $[a_1p_n] + [a_2p_n] + \dots + [a_{2014p_n}] = p_nx_n$. It is obvious that when $n \to \infty$ the sequence $x_n = \frac{[a_1p_n] + [a_2p_n] + \dots + [a_{2014p_n}]}{p_n}$ converges to $a_1 + a_2 + \dots + a_{2014}$ and because all terms of the sequence $(x_n)_{n \ge N}$ are integers, there exists $P$ such that $\forall n \ge P : x_n = a_1+a_2+\dots+a_{2014}$. Since the latter is equivalent to $\forall n \ge P : \{a_1p_n\} + \{a_2p_n\} + \dots + \{a_{2014p_n}\} = 0$, we conclude that $\forall n \ge P : a_ip_n,\ i = 1, 2, \dots, 2014$ are integers. It leads to that $a_i,\ i = 1, 2, \dots, 2014$ are integers and but this is a contradiction.