Bulgarian Mathematical Competitions

Solution:

We first prove that for any odd $n \geq 3$ there are $n \times n$ tables that are not "good". Consider an arbitrary $n \times n$ table and denote by $P_{i}$, $i=1,2, \ldots, n$, the product of the numbers in the $i$-th row at the second last step. Then $P_{1} P_{3} = P_{2} P_{4} = \cdots = P_{n-1} P_{1} = P_{n} P_{2} = 1$ and since $n$ is odd, it follows that $P_{1} = P_{2} = \cdots = P_{n}$. This argument shows that the row products in the initial table must be equal. Therefore any table which has not this property is not "good".

We now consider a table of order $n = 2^{k} m$, where $m$ is an odd number and $k \geq 1$. After the first two steps the number in the position $(i, j)$ becomes equal to the product of the numbers in the positions $(i-2, j)$, $(i, j-2)$, $(i, j+2)$ and $(i+2, j)$. Therefore the resulting table after every even step can be obtained by applying the operation on the following four tables of order $2^{k-1} m$:
- the table of all $(i, j)$ with $i \equiv j \equiv 0 \pmod{2}$;
- the table of all $(i, j)$ with $i \equiv 0 \pmod{2}$, $j \equiv 1 \pmod{2}$;
- the table of all $(i, j)$ with $i \equiv 1 \pmod{2}$, $j \equiv 0 \pmod{2}$;
- the table of all $(i, j)$ with $i \equiv j \equiv 1 \pmod{2}$.

Now it follows by induction that the number $n = 2^{k} m$ has the required property if and only if the number $2^{k-1} m$ does. It is also easy to see that every table of order $2$ is "good".

Therefore the required $n$ are $n = 2^{k}$, where $k$ is a positive integer.