3. Let $\left(2+x-2 x^{2}\right)^{1005}=\sum_{k=0}^{2010} a_{k} x^{k}$. Then
$$
\begin{array}{l}
a_{1}+3 a_{3}+5 a_{5}+\cdots+2009 a_{2009} \\
=
\end{array}
$$
3. 1005.

Let $f(x)=\left(2+x-2 x^{2}\right)^{1005}$. Then $f^{\prime}(x)=1005\left(2+x-2 x^{2}\right)^{1004}(1-4 x)$. Therefore, $f^{\prime}(1)=1005 \times(-3)$, $f^{\prime}(-1)=1005 \times 5$.
Given $f(x)=\sum_{k=0}^{2010} a_{k} x^{k}$, we have
$$
\begin{array}{l}
f^{\prime}(x)=\sum_{k=1}^{2010} k a_{k} x^{k-1} . \\
\text { Hence } a_{1}+3 a_{3}+5 a_{5}+\cdots+2009 a_{2000} \\
=\frac{1}{2}\left(f^{\prime}(1)+f^{\prime}(-1)\right)=1005 .
\end{array}
$$