9. In the sequence $\left\{a_{n}\right\}$, $a_{1}=-1, a_{2}=1, a_{3}=-2$. If for all $n \in \mathbf{N}_{+}$, $a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+$ $a_{n+2}+a_{n+3}$, and $a_{n+1} a_{n+2} a_{n+3} \neq 1$, then the sum of the first 4321 terms $S_{4321}$ of the sequence is
9. -4321
When $n=1$, from $a_{1} a_{2} a_{3} a_{4}=a_{1}+a_{2}+a_{3}+a_{4}$, we get $2 a_{4}=-2+a_{4}$, so $a_{4}=-2$; when $n=2$, from $a_{2} a_{3} a_{4} a_{5}=a_{2}+a_{3}+a_{4}+a_{5}$, we get $4 a_{5}=-3+a_{5}$, so $a_{5}=-1$; when $n=3$, from $a_{3} a_{4} a_{5} a_{6}=a_{3}+a_{4}+a_{5}+a_{6}$, we get $-4 a_{6}=-5+a_{6}$, so $a_{6}=1$; .....

From this, we can see that $\left\{a_{n}\right\}$ is a periodic sequence with the smallest positive period being 4. Since the sum of the first 4 terms is -4, the sum of the first 4321 terms $S_{4321}=1080 \times(-4)-1=-4321$. Therefore, the answer should be -4321.