6. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=a_{2}=1, a_{3}=2, \\
a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}(n \geqslant 1),
\end{array}
$$

and $a_{n} a_{n+1} a_{n+2} \neq 1$ for all positive integers $n$. Then the value of $S_{100}=\sum_{k=1}^{100} a_{k}$ is $(\quad)$.
(A) 225
(B) 200
(C) 300
(D) 150
6. B.

For any $n \geqslant 1$, we have
$$
\begin{array}{l}
a_{n} a_{n+1} a_{n+2} a_{n+3}=a_{n}+a_{n+1}+a_{n+2}+a_{n+3}, \\
a_{n+1} a_{n+2} a_{n+3} a_{n+4} \\
=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4} .
\end{array}
$$
Subtracting (2) from (1) gives
$$
\begin{array}{l}
a_{n+1} a_{n+2} a_{n+3}\left(a_{n}-a_{n+4}\right)=a_{n}-a_{n+4}, \\
\left(a_{n}-a_{n+4}\right)\left(1-a_{n+1} a_{n+2} a_{n+3}\right)=0 .
\end{array}
$$

Since $a_{n+1} a_{n+2} a_{n+3} \neq 1$, it follows that $a_{n}=a_{n+4}$.
Thus, $\left\{a_{n}\right\}$ is a sequence with a period of 4.
Given $a_{1}=a_{2}=1, a_{3}=2, a_{4}=4$, we have
$$
S_{100}=\sum_{k=1}^{100} a_{k}=25\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=25 \times 8=200 .
$$