Let its two roots are $x_1$ and $x_2$. Then we have $x_1+x_2 > 0$ as well as $x_1x_2 > 0$. By Vieta's theorem $$ \left\{ \begin{array}{cl} (m-2n) & > 0\\ \frac{1}{4}mn & > 0 \end{array} \right. \implies m > 2n > 0 $$ Meanwhile, because the two roots are equal, we have $$(m-2n)^2 - mn = 0 \implies (m-4n)(m-n)=0$$ Therefore $$m-4n=0\implies \frac{m}{n}=\boxed{4}$$