Argentina_2018

Here is one way to construct such a table. Start by satisfying only the second part of the condition, that the products by columns are equal. This holds for the first table $T$ below. It has dimensions $4\times 4$ and all column products equal $120$. The second table has an additional column in which the row sums of $T$ are shown.

$$T=\begin{array}{cccc}4& 1& 2& 8\\ 3& 2& 5& 1\\ 2& 10& 3& 5\\ 5& 6& 4& 3\end{array}\to \begin{array}{ccccc}528& 132& 264& 1056& 1980\\ 540& 360& 900& 180& 1980\\ 198& 990& 297& 495& 1980\\ 550& 660& 440& 330& 1980\end{array}$$

Note that the equality of the column products is preserved if a row of $T$ is multiplied by any number. We use this observation to equalize the different row sums by choosing an appropriate integer for each row. To this end consider the least common multiple $1980$ of the row sums, $15$, $11$, $20$, $18$.

Multiply rows 1, 2, 3, 4 respectively by $\frac{1980}{15}=132$, $\frac{1980}{11}=180$, $\frac{1980}{20}=99$, $\frac{1980}{18}=110$. The new table has row sums equal to $1980$ and equal column products (their common value is the large number $31049568000$). It remains to observe that all new numbers are different.