**Now duplicate of Magic $\mathbb{Z}\times\mathbb{Z}$-square where it has an answer.**

Is there an injective map $j:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$ satisfying: For every $z\in \mathbb{Z}$ we have $$\lim_{N\to \infty}\sum_{k=-N}^Nj(k,z) = 0 = \lim_{N\to \infty}\sum_{k=-N}^Nj(z,k)\text{ ?}$$

That is, for every $z\in \mathbb{Z}$ there is $N_0=N_0(z)\in \mathbb{N}$, such that for every integer $N\geq N_0$ we have $$\sum_{k=-N}^Nj(k,z) = 0 = \sum_{k=-N}^Nj(z,k).$$

**Note.**
Martin Sleziak has answered this question in another thread: https://mathoverflow.net/a/344049/8628 (no answers could be posted here since this question is on hold).