Mongolian Mathematical Olympiad

Let's form from every triple $\{a,b,c\}$ a pair, namely $\{a,b\}$. By the given condition, different triples must correspond to different pairs. The pair $\{a,b\}$ represents pairs $\{a,c\}$, $\{b,c\}$. Since total number of pairs is $\frac{n(n-1)}{2}$, number of representing pairs not greater than $\frac{1}{3}\cdot \frac{n(n-1)}{2}$. In other words $k\le \frac{n(n-1)}{6}$.

If one forms a plan satisfying given condition of at least $\frac{n(n-3)}{6}$ working days then $k\ge \frac{n(n-3)}{6}$.

Let enumerate students $\{1,2,...,n\}$. It is obvious that all triples $a<b<c$ with $n\mid a+b+c$ satisfies the condition. There are $n$ choices of $a$ and at least $n-3$ choices of $b$. Therefore we have at least $\frac{n(n-3)}{3!}$ choices of triples with the property $n\mid a+b+c$, $a<b<c$.

Note. Given problem implies ${\lim }_{n\to \infty }\frac{k}{n^2}=\frac{1}{6}$. In other words the asymptotic estimation $k\sim \frac{1}{6}{n}^2$, $n\to \infty$ holds.