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).

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    $\begingroup$ Probably it might be useful to clarify in what meaning $\sum\limits_{k\in\mathbb Z} x_k$ is meant. Do you mean $\lim\limits_{N\to\infty} \sum\limits_{k=-N}^{k=N} x_k$? Do you mean "limit of sums of finite sets" - which is the way in which sums over arbitrary sets are often defined? (Although if it is the latter, the proof that this is impossible seems rather easy.) Do you mean something completely else? $\endgroup$ – Martin Sleziak Oct 14 '19 at 9:23
  • $\begingroup$ Right - thanks @MartinSleziak for making me aware of this inambiguity! I have just corrected it. $\endgroup$ – Dominic van der Zypen Oct 14 '19 at 10:47
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    $\begingroup$ I am not sure why this problem was down-voted or why it was put on hold. There is at least one paper in the literature on infinite magic squares (different from this generalization): Antoine Salomon, "Infinite sized magic square," the American Mathematical Monthly (January 2014). hal.archives-ouvertes.fr/hal-01401844v4/document There is also a discussion elsewhere on this site about infinite magic squares. mathoverflow.net/questions/53420/… If this problem is somehow trivial, then someone can provide the solution. $\endgroup$ – Tri Oct 15 '19 at 15:06
  • $\begingroup$ Thanks @Tri! Seeing the downvotes, I assumed there must be a trivial answer to my question - but please can anyone give a hint to a trivial example or a trivial reason why such a $\mathbb{Z}\times\mathbb{Z}$-square can not exist? $\endgroup$ – Dominic van der Zypen Oct 16 '19 at 6:52
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    $\begingroup$ @DominicvanderZypen I've copied the question here to the other thread. I suggest you delete this one since it doesn't need to be in 2 threads. (For memory, the question was initially downvoted and closed because the sum was not specified to be from $-N$ to $+N$, so the convergence was obviously impossible). $\endgroup$ – YCor Oct 17 '19 at 22:49