4. 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 $a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{1}+$ $a_{n+1}+a_{n+2}+a_{n+3}$, then the value of $a_{1}+a_{2}+\cdots+a_{100}$ is $\qquad$.
$$
\begin{aligned}
5 . \because \quad a_{1} & =a_{2}=1, \\
a_{3} & =2,
\end{aligned}
$$

and $a_{1} a_{2} a_{3} a_{4}=a_{1}+a_{2}+a_{3}+a_{4}$,
$$
\therefore \quad a_{4}=4 \text {. }
$$

From the given conditions,
$$
\begin{aligned}
& a_{n} a_{n+1} a_{n+2} a_{n+3} \\
= & a_{n}+a_{n+1}+a_{n+2}+a_{n+3}, \\
& a_{n+1} a_{n+2} a_{n+3} a_{n+4} \\
= & a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4} .
\end{aligned}
$$

Subtracting the above two equations, we get
that is,
$$
\begin{array}{c}
a_{n+1} a_{n+2} a_{n+3}\left(a_{n}-a_{n+4}\right)=a_{n}-a_{n+4}, \\
\left(a_{n}-a_{n+4}\right)\left(a_{n+1} a_{n+2} a_{n+3}-1\right)=0
\end{array}
$$

According to the given conditions $a_{n+1} a_{n+2} a_{n+3} \neq 1$, hence $a_{n+4}=a_{n}$.
Therefore,
$$
\begin{aligned}
\sum_{i=1}^{100} a_{k} & =\frac{100}{4}\left(a_{1}+a_{2}+a_{3}+a_{4}\right) \\
& =200 .
\end{aligned}
$$