jmc

Since $x-3$ is a multiple of $7$, we know that $x\equiv 3(\mathrm{mod}7)$.

Since $y+3$ is a multiple of $7$, we know that $y\equiv -3(\mathrm{mod}7)$.

Therefore, $$\begin{array}{rl}{x}^2+xy+y^2+n& \equiv (3{)}^2+(3)(-3)+(-3{)}^2+n\\ & \equiv 9-9+9+n\\ & \equiv 9+n(\mathrm{mod}7).\end{array}$$In other words, $9+n$ is a multiple of $7$. The smallest positive $n$ for which this is true is $n=\boxed{5}$.