Let $a$ and $b$ be real numbers such that $\frac{1}{a^2} +\frac{3}{b^2} = 2018a$ and $\frac{3}{a^2} +\frac{1}{b^2} = 290b$. Then $\frac{ab}{b-a }= \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Let's think step by step and output the final answer within \boxed{}.
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