$$
\begin{array}{l}
\text { 6. Given } x, y \in \mathbf{R}, \\
z_{1}=x+(y+2) \mathrm{i}, \\
z_{2}=(x-2)+y \mathrm{i}, \\
\left|z_{1}\right|+\left|z_{2}\right|=4 .
\end{array}
$$

Then the maximum value of $|x+y|$ is
6. $2 \sqrt{2}$.

From the problem, we know that the point $(x, y)$ lies on an ellipse with foci at $(0,-2)$ and $(2,0)$, and a major axis length of 4.

Notice that the center of the ellipse is $(1,-1)$, and the slope of the major axis is 1.

Therefore, when $(x, y)$ is at the vertex of the major axis, $|x+y|$ is maximized. At this point, the vertices of the major axis are $(\sqrt{2}+1, \sqrt{2}-1)$ and $(-\sqrt{2}+1, -\sqrt{2}-1)$.
Thus, the maximum value is $2 \sqrt{2}$.