jmc

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 $$\begin{array}{rl}& (x-5)(x-10)(x-25)+(x-1)(x-10)(x-25)\\ & +(x-1)(x-5)(x-25)+(x-1)(x-5)(x-10)=2(x-1)(x-5)(x-10)(x-25).\end{array}$$(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 $\frac{86}{2}=\boxed{43}.$