We seek to use Vieta's formulas. To be able to apply the formulas, we multiply both sides by $ (x-1)(x-5)(x-10)(x-25) $ to eliminate the fractions. This gives \[\begin{aligned}&\quad (x-5)(x-10)(x-25) + (x-1)(x-10)(x-25) \\& + (x-1)(x-5)(x-25) + (x-1)(x-5)(x-10) = 2(x-1)(x-5)(x-10)(x-25).\end{aligned}\](Be careful! We may have introduced one of the roots $ x = 1, 5, 10, 25 $ into this equation when we multiplied by $ (x-1)(x-5)(x-10)(x-25) $. However, note that none of $ x = 1, 5, 10, 25 $ satisfy our new equation, since plugging each one in gives the false equation $ 1=0 $. Therefore, the roots of this new polynomial equation are the same as the roots of the original equation, and we may proceed.) The left-hand side has degree $ 3 $ while the right-hand side has degree $ 4, $ so when we move all the terms to the right-hand side, we will have a $ 4 $th degree polynomial equation. To find the sum of the roots, it suffices to know the coefficients of $ x^4 $ and $ x^3 $. The coefficient of $ x^4 $ on the right-hand side is $ 2, $ while the coefficients of $ x^3 $ on the left-hand and right-hand sides are $ 4 $ and $ 2(-1-5-10-25) = -82, $ respectively. Therefore, when we move all the terms to the right-hand side, the resulting equation will be of the form \[ 0 = 2x^4 - 86x^3 + \cdots\]It follows that the sum of the roots is $ \tfrac{86}{2} = \boxed{43} $.