This problem has been bugging me for a few days, but I now see how to do it. The idea is to fill row n and column n using numbers whose denominators are 2^n times an odd number. Denote this set of numbers by , so that Notice that the elements of form the terms of a divergent series.

For convenience, labels the rows and the columns with indices starting at 0 rather than 1. So the first row is row 0. Fill that row as follows. Take the reciprocals of all the odd numbers in turn. For each such number, put it in row 0 provided that the sum of all the numbers in that row stays below 2. Row 0 will then look like this:

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Now put all the remaining reciprocals of odd numbers into column 0. That column will then contain the elements

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Next, fill row 1 and column 1 in a similar way, using the set (reciprocals of all the odd multiples of 2). Row 1 already contains the element , so we want the sum of its remaining elements to be . Taking each element of in turn, put it in row 1 provided that the sum of all the numbers in that row stays below 2, and put it in column 1 otherwise.

Now proceed inductively. Suppose that rows and columns 0 to n–1 have already been filled, and fill row n and column n as follows. The first n–1 elements of row n will already have been filled. But their sum will be less than 2, because each element in column k (for ) is in and is therefore less than . So take each element of in turn, put it in row n provided that the sum of all the numbers in that row stays below 2, and put it in column n otherwise.

That way, the sum in each row will be 2, the sum in each column will be infinite, and each number of the form 1/n will be used exactly once.