3. Given a sequence of numbers $a_{1}, a_{2}, \cdots, a_{2006}$, where $a_{1}$ $=1$, and the sum of each pair of consecutive terms is 3. Then $a_{1}-a_{2}+$ $a_{3}-a_{4}+\cdots+a_{2003}-a_{2004}+a_{2005}=$ $\qquad$
3. -1001 .

Given $a_{1}+a_{2}=a_{2}+a_{3}=\cdots=a_{2004}+a_{2000}=3$, and $a_{1}=1$, we can sequentially deduce that
$$
\begin{array}{l}
a_{1}=a_{3}=\cdots=a_{2005}=1, a_{2}=a_{4}=\cdots=a_{2004}=2 . \\
\text { Then } a_{1}-a_{2}+a_{3}-a_{4}+\cdots-a_{2004}+a_{2000} \\
=\left(a_{1}-a_{2}\right)+\left(a_{3}-a_{4}\right)+\cdots+\left(a_{2003}-a_{2004}\right)+a_{2000} \\
=-1002+1=-1001 .
\end{array}
$$