Trivial problem: My solution : If $n\ge 2016$ then assume that there are $n$ distinct positive integers that they satisfy the condition. Then let $S$ product of the numbers and assume that $x$ is minimum number among the $n$ numbers then by the condition $S\mid x^{2016}\longrightarrow x^{2016}<S\le x^{2016}$ contradiction so $n\le 2015$ we show by induction that for any $3\le n \le 2015$ the claim is true for $n=3$ consider $6,18,36$ now assume that the claim is true for $n <2015$ so there are $ n $ distinct natural numbers $a_1, a_2,\cdots , a_n$ that satisfy the condition. Let $ S=a_1a_2\cdots a_n$ now consider the $n+1$ numbers $b_1=S, b_2=Sa_1, \cdots , b_{n+1}=Sa_n$ clearly all of them are distinct and you can easily observe that they satisfy the condition. DONE

andria wrote: Trivial problem: My solution : If $n\ge 2016$ then assume that there are $n$ distinct positive integers that they satisfy the condition. Then let $S$ product of the numbers and assume that $x$ is minimum number among the $n$ numbers then by the condition $S\mid x^{2016}\longrightarrow x^{2016}<S\le x^{2016}$ contradiction so $n\le 2015$ we show by induction that for any $3\le n \le 2015$ the claim is true for $n=3$ consider $6,18,36$ now assume that the claim is true for $n <2015$ so there are $ n $ distinct natural numbers $a_1, a_2,\cdots , a_n$ that satisfy the condition. Let $ S=a_1a_2\cdots a_n$ now consider the $n+1$ numbers $b_1=S, b_2=Sa_1, \cdots , b_{n+1}=Sa_n$ clearly all of them are distinct and you can easily observe that they satisfy the condition. DONE Sorry to bother , but i think you're solution is wrong because you understood the question the wrong way . It says (a_i)^2015|S