https://www.artofproblemsolving.com/community/q1h1366481p7516530

Let $\alpha$ be a root of this polynomial with $$0<\alpha<\frac{1}{2016}.$$Set $Q(x)=x^{\text{deg} P}P\left(\frac{1}{x}\right)$ where $P$ is the given polynomial. Note that $Q(0) \ne 0$ any root $\beta>0$ of $Q$ satisfies $\beta>2016$. Finally, we write $$Q(x)=a_nx^n+\dots+a_1x+a_0,$$and then $$|\beta|^n=\frac{1}{|a_n|}|Q(\beta)-a_n\beta^n| \le 2015\left(1+|\beta|+\dots+|\beta^{n-1}|\right)=2015\left(\frac{|\beta|^n-1}{|\beta|-1}\right)<2015\frac{|\beta|^n}{|\beta|-1} \Longrightarrow |\beta|-1<2015,$$yielding the desired contradiction.