jmc

Since $i^4=1,$ $i^{2004-F_j}=\frac{1}{i^{F_j}}$ depends only on the value of $F_j$ modulo 4.

We compute the first few Fibonacci numbers modulo 4: $$\begin{array}{cc}n& F_n(\mathrm{mod}4)\\ 1& 1\\ 2& 1\\ 3& 2\\ 4& 3\\ 5& 1\\ 6& 0\\ 7& 1\\ 8& 1\end{array}$$Since $F_7\equiv F_1\equiv 1(\mathrm{mod}4)$ and $F_8\equiv F_2\equiv 1(\mathrm{mod}4),$ and each term depends only on the previous two terms, the Fibonacci numbers modulo 4 becomes periodic, with period 6.

Since $2004=334\cdot 6,$ $$\sum _{j=1}^{2004}\frac{1}{i^{F_j}}=334\left(\frac{1}{i}+\frac{1}{i}+\frac{1}{i^2}+\frac{1}{i^3}+\frac{1}{i}+\frac{1}{1}\right)=\boxed{-668i}.$$