jmc

By the binomial theorem, $$\begin{array}{rl}{\left(\frac{1}{2}-x\right)}^{2001}& =(-x{)}^{2001}+(\genfrac{}{}{0ex}{}{2001}{1})\left(\frac{1}{2}\right)(-x{)}^{2000}+\left(\genfrac{}{}{0ex}{}{2001}{2}\right){\left(\frac{1}{2}\right)}^2(-x{)}^{1999}+\cdots \\ & =-x^{2001}+\frac{2001}{2}{x}^{2000}-\frac{2001\cdot 2000}{8}{x}^{1999}+\cdots .\end{array}$$Thus, $$x^{2001}+{\left(\frac{1}{2}-x\right)}^{2001}=\frac{2001}{2}{x}^{2000}-\frac{2001\cdot 2000}{8}{x}^{1999}+\cdots .$$(Note that the $x^{2001}$ terms canceled!) Then by Vieta's formulas, the sum of the roots is $$-\frac{-2001\cdot 2000/8}{2001/2}=\boxed{500}.$$

Another approach is to replace $x$ with $\frac{1}{4}-y$, making the equation ${\left(\frac{1}{4}-y\right)}^{2001}+{\left(\frac{1}{4}+y\right)}^{2001}=0$. Because the left-hand side is symmetric with respect to $y$ and $-y$, any solution to the equation pairs off with another solution to have a sum of $0$. Since $x^{2001}$ pairs off with a term of $-x^{2001}$ from the expansion of the second term in the original equation, this is a degree-$2000$ polynomial equation in disguise, so the sum of its roots is $2000\cdot \frac{1}{4}-0=500$.