6. C

From $\left\{\begin{array}{l}5 x+10 y=30 \\ 2 x-y=3\end{array}\right.$ we get the intersection point $D\left(\frac{12}{5}, \frac{9}{5}\right)$. Substituting into $x+y$ gives: $x^{1}+y^{1}=\frac{21}{5}=4 \frac{1}{5}$, so $x+y \leqslant 4 \frac{1}{5}$. Since $x, y \in N$, we have $x+y \leqslant 4$. When $x=y=2$, $(x+y)_{\max }=4$ (Note: $x+y$ taking 4 may sometimes have more than one set of $x, y$).
6. Given: $\left\{\begin{array}{l}5 x+10 y \leqslant 30 \\ 2 x-y \leqslant 3 \\ x \cdot y \in N\end{array}\right.$ Then the maximum value of $x+y$ is ().
A. 6
B. 5
C. 4
D. 3