jmc

Since $z_1=0$, it follows that $c_0=P(0)=0$. The nonreal zeros of $P$ must occur in conjugate pairs, so ${\sum }_{k=1}^{2004}{b}_k=0$ and ${\sum }_{k=1}^{2004}{a}_k=0$ also. The coefficient $c_{2003}$ is the sum of the zeros of $P$, which is $$\sum _{k=1}^{2004}{z}_k=\sum _{k=1}^{2004}{a}_k+i\sum _{k=1}^{2004}{b}_k=0.$$Finally, since the degree of $P$ is even, at least one of $z_2,\dots ,z_{2004}$ must be real, so at least one of $b_2,\dots ,b_{2004}$ is 0 and consequently $b_2{b}_3\cdots b_{2004}=0$. Thus the quantities in $\text{(A)}$, $\text{(B)}$, $\text{(C)}$, and $\text{(D)}$ must all be 0.

Note that the polynomial $$P(x)=x(x-2)(x-3)\cdots (x-2003)\left(x+\sum _{k=2}^{2003}k\right)$$satisfies the given conditions, and ${\sum }_{k=1}^{2004}{c}_k=P(1)\ne 0$. That means our answer is $\boxed{\text{E}}$.