$(\partial/\partial x + \partial/\partial y)f=0$ iff $f$ is a function of $x-y$. To see it, you can e.g. make the change of variables $u=x+y$, $v=x-y$, and compute that $\partial/\partial x + \partial/\partial y=2\partial/\partial u$. If $f(u)$ is a polynomial of degree $k$ then $f(x-y)$ has your restriction on degrees iff $k\leq n$, so you get dimension $n+1$ (of the space of polynomials in one variable, of degree $\leq n$). Notice, however, that we need characteristic $0$ (I suppose your polynomials have coefficients in some field). Otherwise we might have $(\partial/\partial u) u^n=0$, and the dimension would increase.