4. (1992 National High School League Question) Let the sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{n} \cdots$ satisfy $a_{1}=a_{2}=1, a_{3}=2$ and for any positive integer $n$, it holds that $a_{n} \cdot a_{n+1} \cdot a_{n+2} \neq 1$, and also $a_{n} \cdot a_{n+1} \cdot a_{n+2} \cdot a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+$ $a_{n+3}$, then the value of $a_{1}+a_{2}+\cdots+a_{100}$ is $\qquad$ .
4. 200. Since $a_{n} \cdot a_{n+1} \cdot a_{n+2} \cdot a_{n+3}=a_{1}+a_{n+1}+a_{n+2}+a_{n+3}, a_{n+1} \cdot a_{n+2} \cdot a_{n+3} \cdot a_{n+4}=a_{n+1}+$ $a_{n+2}+a_{n+3}+a_{n+4}$, subtracting the two equations gives $a_{n+1} \cdot a_{n+2} \cdot a_{n+3} \cdot\left(a_{n+4}-a_{n}\right)=a_{n+4}-a_{n}$. Since $a_{n+1} \cdot a_{n+2} \cdot a_{n+3} \neq 1$, it follows that $a_{n+4}=a_{n}$, hence $a_{1}+a_{2}+\cdots+a_{100}=25\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=25(1+1+2+4)=200$. Here $a_{4}=4$ is derived from $a_{1} a_{2} a_{3} a_{4}=a_{1}+a_{2}+a_{3}+a_{4}$ and $a_{1}=a_{2}=1, a_{3}=2$.