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I don't think so (but I haven't checked this argument very thoroughly):

First we claim that any such $f$ has degree two. Clearly the leading term of $f$ cannot be odd, so suppose by contradiction that $f$ has degree at least four. Pick a constant $R$ large enough so that in the region $D_1$ consisting of points satisfying $|x|, |y| \ge R$, there exists a constant $c$ such that $f(x, y) \ge c \text{ min}(|x|, |y|)^4$. (Edit: This isn't always possible, but I think it can be salvaged.) Then it's not hard to see that $\sum_{D_1} \frac{1}{f(x, y)}$ converges. But $D_2 = \mathbb{Z}^2 - D_1$ can be partitioned into $4R - 2$ not necessarily disjoint lines in which one of $x$ or $y$ is fixed. On any of these regions $f$ cannot be linear, so it either grows at least quadratically or is a constant; we can ignore the lines on which $f$ is constant. It follows that $\sum_L \frac{1}{f(x, y)}$ converges for any line $L$ on which $f$ is nonconstant, hence $\sum \frac{1}{f(x, y)}$ converges if we sum over every point of $\mathbb{Z}^2$ except the lines on which $f$ is constant. Since we have only thrown out finitely many of the values of $f$ in this sum, those values cannot contain every positive integer.

But if $f$ is quadratic, it is a constant plus the sum of squares of two polynomials with rational coefficients and there are many integers not representable as the sum of squares of two rational numbers.

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I don't think so (but I haven't checked this argument very thoroughly):

First we claim that any such $f$ has degree two. Clearly the leading term of $f$ cannot be odd, so suppose by contradiction that $f$ has degree at least four. Pick a constant $R$ large enough so that in the region $D_1$ consisting of points satisfying $|x|, |y| \ge R$, there exists a constant $c$ such that $f(x, y) \ge c \text{ min}(|x|, |y|)^4$. Then it's not hard to see that $\sum_{D_1} \frac{1}{f(x, y)}$ converges. But $D_2 = \mathbb{Z}^2 - D_1$ can be partitioned into $4R - 2$ not necessarily disjoint lines in which one of $x$ or $y$ is fixed. On any of these regions $f$ cannot be linear, so it either grows at least quadratically or is a constant; we can ignore the lines on which $f$ is constant. It follows that $\sum_L \frac{1}{f(x, y)}$ converges for any line $L$ on which $f$ is nonconstant, hence $\sum \frac{1}{f(x, y)}$ converges if we sum over every point of $\mathbb{Z}^2$ except the lines on which $f$ is constant. Since we have only thrown out finitely many of the values of $f$ in this sum, those values cannot contain every positive integer.

But if $f$ is quadratic, it is a constant plus the sum of squares of two polynomials with rational coefficients and there are many integers not representable as the sum of squares of two rational numbers.
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