We can let $P(0) = \tfrac{1}{2}$, and then have $P(x) = F(x) + \tfrac{1}{2}$. Then $F(x) + F(1 - x) = 0$, and I have found $F(x) = (1 - x)^{2001} - x^{2001}$ satisfies this. So one such polynomial would be \[\boxed{P(x) = (1 - x)^{2011} - x^{2011} + \tfrac{1}{2}}\]