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 $n = a b$ 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 $\prod_{p^{\alpha} \| n} \sigma\left(p^{\alpha}\right)$.
- 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.