Example 6 (1999 Henan Mathematics Competition Question) 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 $a_{n+1} a_{n+2} \neq 1$. Find $\sum_{i=1}^{2006} a_{i}$.
Solution: $a_{3}=\frac{a_{1}+a_{2}}{a_{1} a_{2}-1}=3, a_{4}=\frac{a_{2}+a_{3}}{a_{2} a_{3}-1}=1$, $a_{5}=\frac{a_{3}+a_{4}}{a_{3} a_{4}-1}=2$, and $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}, a_{n+1} a_{n+2} a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3}$, subtracting these gives $a_{n+3}-a_{n}=a_{n+1} a_{n+2}\left(a_{n+3}-a_{n}\right)$, which means $\left(a_{n+3}-a_{n}\right)\left(a_{n+1} a_{n+2}-1\right)=0$.

Since $a_{n+1} a_{n+2} \neq 1$, it follows that $a_{n+3}-a_{n}=0$, or $a_{n+3}=a_{n}$, so $\left\{a_{n}\right\}$ is a periodic sequence with a period of 3. Since $a_{1}+a_{2}+a_{3}=6, 2006 \equiv 2(\bmod 3)$, thus $\sum_{i=1}^{2006} a_{i}=6 \times 668+1+2=4011$.