6. Given $0 \leqslant m-n \leqslant 1, 2 \leqslant m+n \leqslant 4$. When $m-2 n$ reaches its maximum value, the value of $2019 m+2020 n$ is ( ).
(A) 2019
(B) 2020
(C) 0
(D) -1
6. A.

Construct the equation
$$
x(m-n)+y(m+n)=m-2 n,
$$

i.e., $(x+y) m+(y-x) n=m-2 n$.
By comparing coefficients and solving the system of equations, we get
$$
x=\frac{3}{2}, y=-\frac{1}{2} \text {. }
$$

According to the problem, we have
$$
\begin{array}{l}
0 \leqslant \frac{3}{2}(m-n) \leqslant \frac{3}{2},-2 \leqslant-\frac{1}{2}(m+n) \leqslant-1 \\
\Rightarrow-2 \leqslant m-2 n \leqslant 1 .
\end{array}
$$

Thus, the maximum value of $m-2 n$ is 1, at which point, $n=$ $\frac{m-1}{2}$. Substituting into the inequality conditions, we get
$$
0 \leqslant m \leqslant 1,1 \leqslant m \leqslant 3 \text {. }
$$

Therefore, $m=1, n=0$.
Hence, $2019 m+2020 n=2019$.