Hey, isn't that my birth year?

If $a$ is a root, then so is $\frac{1}{2}-a$ ; plug it in! Thus the average value of the roots is $\frac{1}{4}$ . This being a polynomial of degree 2000 (the $x^{2001}$ term cancels out), the sum of the roots is $2000\times \frac{1}{4}=\boxed{500}$

Expand $\left(-x+\dfrac{1}{2} \right )^{2001}$ using Binomial Expansion : $\cancel{x^{2001}} + \left(\cancel{-\binom{2001}{0}x^{2001}}+\binom{2001}{1}x^{2000}\times \dfrac 12 -\binom{2001}{2}x^{1999}\times \dfrac 14\cdots\right)=0$ $\implies\binom{2001}{1}x^{2000}\times \dfrac 12 -\binom{2001}{2}x^{1999}\times \dfrac 14\cdots=0$ By Vieta's formula . sum of roots of this equation are given by:-

$\dfrac{\binom{2001}{2}\times \dfrac 14}{\binom{2001}{1}\times \dfrac 12}=\boxed{500}$

See also this basically identical question and associated solutions: Will you really find roots?

Can any one explain why this is level 4