Magic square on an infinite lattice This question came to me while reading the discussion of magic square in the complex plane with equal integrals along every horizontal, vertical and diagonal "magic square in the complex plane with equal integrals along every horizontal, vertical and diagonal"
That question ran into trouble when people pointed out the difficulties involved in just making sense of the question, e.g., what kind of function, what kind of integral, etc. So I made up something halfway between the finite-discrete magic square (that is, the usual kind) and the continuous analogue raised in Question 53352: 
What can you say about complex numbers $a_{i,j}$ such that for all $i$ and $j$ the sums $$\sum_{i=-\infty}^{\infty}a_{i,j}=\sum_{j=-\infty}^{\infty}a_{i,j}\lt\infty$$ That is, the sums along vertical and horizontal lines all converge and are all equal. 
If the doubly-infinite sums are a worry, just insist that all the numbers be non-negative reals (alternatively, replace the lower limits of summation with zero). 
Splitting into real and imaginary parts, we see that we may assume the numbers are real. Other than that, I haven't done much thinking about it. 
I realize the number theory tag is not quite appropriate; I'm trying to comply with the direction, "Please try use at least one tag corresponding to an arXiv subject area." 
 A: Here are two elementary results about those magic square arrays.
Claim: The vector subspace of collections with finitely many nonzero entries has abasis consisting of the magic squares with $4$ nonzero entries arranged as 
$$\begin{array}{cc}-1 & 1 \\\1 & -1\end{array}$$.
Proof: You can always add a multiple of such a magic square to eliminate the rightmost nonzero entry in the bottom row without extending the nonzero entries to the top or left.
Claim: Consider arrays with all indices positive. Any arbitrarily chosen convergent first row and first column with the same sums can be extended to a magic square.
Proof: This can be done so that the rest of the array is $0$ except for the diagonal and one off-diagonal. For $i=2,3,4, ...$ choose $a_{i,i}$ to make the $i$th column have the required sum. Then choose $a_{i+1,i}$ to make the $i$th row have the required sum.
I think infinite arrays are a little too flexible. You can specify large portions of the array arbitrarily, such as the diagonal and everything below it (subject to convergence), and you can complete the array to a magic square.
