$ \sqrt {m} + \frac {2005}{\sqrt {n}} = 2006$ $ \frac {2005}{\sqrt {n}} = 2006 - \sqrt {m}$ $ \dfrac{2005^2}{n} = 2006^2 + m - 4012\sqrt {m}$ Since the left side is rational, the right side must also be rational. Thus, $ \sqrt {m}$ must be rational. Since $ m$ is an integer, this means that $ m$ must be a integral perfect square. So let $ m = p^2$ for some positive integer $ p$. Plugging this into the original equation yields: $ p + \frac {2005}{\sqrt {n}} = 2006$. Since $ p$ and $ 2006$ are integers, $ \frac {2005}{\sqrt {n}}$ must also be an integer. Thus, $ n$ must also be an integer perfect square. So let $ n = q^2$ for some positive integer $ p$. This yields the equation: $ p + \dfrac{2005}{q} = 2006$. Since $ \dfrac{2005}{q}$ must be an integer, $ q \mid 2005 = 5 \cdot 401$. Thus, $ q = 1, 5, 401, 2005$. These yield $ p = 1, 1605, 2001, 2005$ respectively. Therefore, the solutions are: $ (m,n) = (1,1); (1605^2,25); (2001^2,401^2); (2005^2,2005^2)$.