3. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=2$, $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}$, and $a_{n+1} a_{n+2} \neq$ 1. Then $a_{1}+a_{2}+\cdots+a_{2004}=$ $\qquad$
3.4008.

Substituting $a_{1}=1, a_{2}=2$ into $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+$ $a_{n+2}$, we get $a_{3}=3$.
$$
\begin{array}{l}
\text { From } a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}, \\
a_{n+1} a_{n+2} a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3},
\end{array}
$$

Subtracting the two equations, we get $\left(a_{n+3}-a_{n}\right)\left(a_{n+1} a_{n+2}-1\right)=0$.
Since $a_{n+1} a_{n+2} \neq 1$, it follows that $a_{n+3}=a_{n}$.
Therefore, the sequence $\left\{a_{n}\right\}$ is a periodic sequence.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{2000} \\
=\left(a_{1}+a_{2}+a_{3}\right)+\left(a_{4}+a_{5}+a_{6}\right)+\cdots+ \\
\left(a_{2002}+a_{2003}+a_{2004}\right) \\
=(1+2+3) \times 668=4008 \text {. } \\
\end{array}
$$