Mongolian Mathematical Olympiad Round 3

Let us say a pair of students is friends if they have sat in one group for at least a month.
Now we prove that four months is not enough. First note that for integers $x \ge 0$, $y \ge 0$, $z \ge 0$ such that $x + y + z = 10$, we have
$$f(x, y, z) = \binom{x}{2} + \binom{y}{2} + \binom{z}{2} \ge 12.$$
Indeed, by the arithmetic-quadratic mean inequality, we have
$$f(x, y, z) = \frac{1}{2}(x^2 + y^2 + z^2) - \frac{1}{2}(x + y + z) \ge \frac{1}{6}(x + y + z)^2 - \frac{1}{2}(x + y + z) = 11 + \frac{2}{3}$$
and since $f(x, y, z)$ is an integer, we have $f(x, y, z) \ge 12$.

Let $S$ denote the set of students in the class. Then $|S| = 30$. Let $X, Y, Z$ denote the groups of the first month. Here $X, Y, Z \subseteq S$ and $|X| = |Y| = |Z| = 10$. Suppose that for the next three months, the students sat in groups as follows:
$$T_1, T_2, T_3 \quad T_4, T_5, T_6 \quad T_7, T_8, T_9.$$
For a group $T \subseteq S$, each pair in $X \cap T$ is already friends and similarly for $Y \cap T$ and $Z \cap T$. Hence $T$ creates at most
$$\binom{|T|}{2} - \binom{|X \cap T|}{2} - \binom{|Y \cap T|}{2} - \binom{|Z \cap T|}{2} \le 45 - 12 = 33$$
new friends. Here $|T| = 10$ and $\binom{10}{2} = 45$. It follows that in four months at most $3 \cdot 45 + 9 \cdot 33 = 432$ friends are created. There are $\binom{30}{2} = 435$ pairs in total, thus four months is not enough to make every pair friends.

Five months is enough. Divide the students into six groups $A, B, C, D, E, F$ of five students each and sit as follows:
$$\begin{vmatrix} 1 & AB & CD & EF \\ 2 & AC & BE & DF \\ 3 & AD & BF & CE \\ 4 & AE & BD & CF \\ 5 & AF & BC & DE \end{vmatrix}$$