1. $3 y^{2}+3 x y+x+y=2019 \Leftrightarrow(x+y)(3 y+1)=2019$. Since $2019=1 \cdot 2019=3 \cdot 673$ and $3 y+1 \equiv 1(\bmod 3)$, the possible cases are $3 y+1=1$, or $3 y+1=673$. From which $(x, y)=(-221,224)$, or $(x, y)=(2019,0)$.
Answer: $(x, y)=(-221,224),(2019,0)$. Answer to variant: 2-2: $(x, y)=(-224,221)$, $(0,-2019)$.