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

and the sum of the first 2019 terms of the sequence $\left\{a_{n}-n\right\}$ is 2019. Then the value of $a_{2020}$ is $\qquad$ .
6.1.
$$
\begin{array}{l} 
\text { From } a_{n+1}+(-1)^{n} a_{n}=2 n-1 \\
\Rightarrow\left\{\begin{array}{l}
a_{2 n+1}+a_{2 n}=4 n-1, \\
a_{2 n}-a_{2 n-1}=4 n-3, \\
a_{2 n+2}-a_{2 n+1}=4 n+1
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a_{2 n+1}+a_{2 n-1}=2, \\
a_{2 n+2}+a_{2 n}=8 n
\end{array}\right. \\
\Rightarrow a_{2 n-1}+a_{2 n}+a_{2 n+1}+a_{2 n+2}=8 n+2 \\
\Rightarrow\left(a_{2 n-1}-(2 n-1)\right)+\left(a_{2 n}-2 n\right)+a_{2 n+1}- \\
(2 n+1)+\left(a_{2 n+2}-(2 n+2)\right)=0 .
\end{array}
$$
Let $b_{n}=a_{n}-n$. Then
$$
b_{2 n-1}+b_{2 n}+b_{2 n+1}+b_{2 n+2}=0 \text {. }
$$

Hence, the sum of the first 2019 terms of $\left\{a_{n}-n\right\}$ is
$$
\begin{array}{l} 
b_{1}+b_{2}+b_{3}+b_{4}+\cdots+b_{2018}+b_{2019} \\
=\left(b_{1}+b_{2}+b_{3}+b_{4}\right)+\left(b_{5}+b_{6}+b_{7}+b_{8}\right)+ \\
\cdots+\left(b_{2013}+b_{2014}+b_{2015}+b_{2016}\right)+ \\
b_{2017}+b_{2018}+b_{2019} \\
= b_{2017}+b_{2018}+b_{2019}=-b_{2020} \\
= 2010-a_{2020}=2019 \\
\Rightarrow a_{2020}=1 .
\end{array}
$$