My solution: It is easy to see that : $2015m<2014m+1007$ since $1\leqslant m\leqslant 1006$ $\Longrightarrow$ $\dfrac{2015}{1007}m<2m+1$ ....$(1)$ It is easy to see that : $2016m+1008<2015(m+1)$ since $1\leqslant m\leqslant 1006$ $\Longrightarrow$ $2m+1<\dfrac{2015}{1008}(m+1)$....$(2)$ By $(1)$ and $(2)$ : $\dfrac{2015}{1007}m<2m+1<\dfrac{2015}{1008}(m+1)$ for every integer $m$ with $1\leqslant m\leqslant 1006$ then $q=2015$ meets the condition . Now we will prove that is the least , for the integer $m=1006$ exist an integer $n$ such that : $\dfrac{1006}{1007}q<n<\dfrac{1007}{1008}q$ . If $q<1008$ then $q-1\leqslant \dfrac{1006}{1007}q<\dfrac{1007}{1008}q<q$ which it is a contradiction. If $q>1007$ $\Longrightarrow$ $\dfrac{1007}{1008}q\leqslant q-1$ $\Longrightarrow$ $\dfrac{1006}{1007}q<q-2<\dfrac{1007}{1008}q\leqslant q-1$ $\Longrightarrow$ $q>2014$ this completes the proof .

q = 1007 x 1008 - 1. This is the minimum q for m = 1006, and fulfills the conditions of the prob also for every m smaller than 1006.