$$
\begin{aligned}
& 3 . \mathrm{C}_{2004}^{0}-\mathrm{C}_{2004}^{2}+\mathrm{C}_{2004}^{4}-\mathrm{C}_{2004}^{6}+\cdots-\mathrm{C}_{2004}^{2002} \\
+ & \mathrm{C}_{2004}^{200}=
\end{aligned}
$$
3. $-2^{1002}$

In the binomial expansion
$$
(1+x)^{n}=\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{1} x+\mathrm{C}_{n}^{2} x^{2}+\cdots+\mathrm{C}_{n}^{n} x^{n}
$$

if we take $n=2004, x=\mathrm{i}$, we get
$$
\begin{array}{l}
(1+\mathrm{i})^{2004} \\
=\mathrm{C}_{2004}^{0}-\mathrm{C}_{2004}^{2}+\mathrm{C}_{2004}^{4}-\mathrm{C}_{2004}^{6}+\cdots-\mathrm{C}_{2004}^{2004}+\mathrm{C}_{2004}^{2004}+ \\
\mathrm{i}\left(\mathrm{C}_{2004}^{1}-\mathrm{C}_{2004}^{3}+\mathrm{C}_{2004}^{5}-\mathrm{C}_{2004}^{7}+\cdots+\mathrm{C}_{2004}^{2004}-\mathrm{C}_{2004}^{2008}\right) .
\end{array}
$$

Also, $1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$, so
$$
\begin{array}{l}
(1+i)^{2004}=2^{1002}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{2004} \\
=2^{1002}(\cos 501 \pi+i \sin 501 \pi)=-2^{1002} .
\end{array}
$$

Therefore, $\mathrm{C}_{2004}^{0}-\mathrm{C}_{2004}^{2}+\mathrm{C}_{2004}^{4}-\mathrm{C}_{2004}^{6}+\cdots-\mathrm{C}_{2004}^{2000}+\mathrm{C}_{2004}^{2004}$ $=-2^{1002}$.
Note: We can also derive
$$
C_{204}^{1}-C_{204}^{3}+C_{204}^{5}-C_{2004}^{7}+\cdots+C_{2044}^{2001}-C_{204}^{208}=0 .
$$