11. Let the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfy $a_{1}=a_{2}=1, a_{3}=2$, and for any natural number $n$, $a_{n} a_{n+1} a_{n+2} \neq 1$, and $a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}$, then the value of $a_{1}+a_{2}+\cdots+a_{100}$ is $\qquad$ .
$$
\begin{array}{l}
\text { According to the problem, }\left\{\begin{array}{l}
a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}, \\
a_{n+1} a_{n+2} a_{n+3} a_{n+4}=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}
\end{array}\right. \\
\Rightarrow a_{n+1} a_{n+2} a_{n+3}\left(a_{n+4}-a_{n}\right)=a_{n+4}-a_{n} \Rightarrow a_{n+4}=a_{n}\left(n \in \mathbf{N}^{*}\right) . \\
\text { Also, } a_{1} a_{2} a_{3} a_{4}=a_{1}+a_{2}+a_{3}+a_{4} \Rightarrow 2 a_{4}=4+a_{4} \Rightarrow a_{4}=4 \Rightarrow a_{1}+a_{2}+a_{3}+a_{4}=8 \\
\Rightarrow a_{1}+a_{2}+\cdots+a_{100}=25\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=25 \cdot 8=200 .
\end{array}
$$