Team selection tests for IMO 2018

Denote $\sigma (n),\tau (n)$ as the sum of divisors and number of divisors of $n$, respectively. Consider a row with assigned number is $a$, and suppose that there are $b$ cells on that row are filled by $a$ then $n=ab$ or $a$ is a divisor of $n$. By condition ii), the number of rows of the table is exactly the number of positive divisors of $n$. Number of nonzero on each row is $\frac{n}{a}$, thus the number of zero on each row is $n-\frac{n}{a}$. Number of zero on table is $$\sum _{a\mid n}\left(n-\frac{n}{a}\right)=\sum _{a\mid n}(n-a)=n\cdot \tau (n)-\sum _{a\mid n}a=n\cdot \tau (n)-\sigma (n).$$ This number is odd and $n$ is even which imply that $\sigma (n)$ is odd. Put $n=2^k\cdot q$ with $q$ is odd number then $q$ is also the maximum odd number on the table. The sum of divisor $\sigma (n)$ of $n$ is of form ${\prod }_{p^{\alpha }\Vert n}\sigma \left(p^{\alpha }\right)$. - If $p=2$ then $\sigma \left(2^k\right)$ is an odd number. - If $p>2$ then $\sigma \left(p^{\alpha }\right)=1+p+\cdots +p^{\alpha }$ must be odd, which implies that $\alpha$ is even. From this, we can conclude that all odd prime divisors of $n$ have even exponent, thus $q$ is a perfect square.