Answer: 802

Solution:

Let the original series have first term $a$ and ratio $r,$ so $\frac{a}{1-r} = 2005.$ The squared series is geometric with first term $a^2$ and ratio $r^2,$ so $$\begin{aligned} \frac{a^2}{1-r^2} &= \frac{a}{1-r} \cdot \frac{a}{1+r} \\ &= 2005 \cdot \frac{a}{1+r} \\ &= 10 \cdot 2005, \end{aligned}$$ which gives $\frac{a}{1+r} = 10.$

Dividing the two equations, $\frac{1+r}{1-r} = \frac{2005}{10},$ so $2(1+r) = 401(1-r),$ giving $403r = 399$ and $r = \frac{399}{403}.$ Since $399 = 3 \cdot 7 \cdot 19$ and $403 = 13 \cdot 31,$ the fraction is in lowest terms, and $m + n = 399 + 403 = 802.$