6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n-1}=a_{n}+a_{n-2}(n \geqslant 3) \text {, }
$$

let the sum of the first $n$ terms be $S_{n}$. If $S_{2018}=2017, S_{2019}=$ 2018 , then $S_{20200}=$ $\qquad$
6. 1010 .

Given $a_{n-1}=a_{n}+a_{n-2}(n \geqslant 3)$
$$
\begin{aligned}
\Rightarrow & a_{n}=a_{n-1}-a_{n-2}=a_{n-2}-a_{n-3}-a_{n-2} \\
& =-a_{n-3}=-\left(-a_{n-6}\right)=a_{n-6}
\end{aligned}
$$
$\Rightarrow\left\{a_{n}\right\}$ is a periodic sequence, with a period of 6
$$
\begin{aligned}
\Rightarrow & S_{n}=a_{n}+a_{n-1}+a_{n-2}+\cdots+a_{1} \\
= & \left(a_{n-1}-a_{n-2}\right)+\left(a_{n-2}-a_{n-3}\right)+\cdots+ \\
& \left(a_{2}-a_{1}\right)+a_{2}+a_{1} \\
= & a_{n-1}+a_{2} \\
\Rightarrow & S_{2018}=a_{2017}+a_{2}=a_{1}+a_{2}=2017, \\
& S_{2019}=a_{2018}+a_{2}=a_{2}+a_{2}=2018 \\
\Rightarrow & a_{1}=1008, a_{2}=1009 \\
\Rightarrow & S_{2020}=a_{2019}+a_{2}=a_{3}+a_{2} \\
& =a_{2}-a_{1}+a_{2}=1010 .
\end{aligned}
$$