SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Firstly, we shall prove that if $A,B$ are friend then the sum of friends of each one does not exceed $300$. Indeed, Suppose that $A$ has $a\le N$ friends and $B$ has $b\le N$ friends. Note that $A$ and $B$ cannot have any common friend; otherwise, take $C$ as a friend of $A,B$ and then the triple ($A,B,C$) does not satisfy the given condition. Thus $$(a-1)+(b-1)\le 298\text{ and }a+b\le 300.$$ Suppose that $N\ge 201$ and take some student $X$ who has $201$ friends. Take $Y$ who has at least $100$ friends, then as the remark above, $X$ and $Y$ are not friends. Note that we always can find $100$ distinct students and each of them has exactly $100,101,\dots ,200$ friends, and they are not friends of $X$. So the number of friends of $X$ is less than $200$, contradiction. Therefore, $N\le 200$.

We can give an example as follows. Divide students into groups $A=\left\{A_1,A_2,\dots ,A_{200}\right\}$ and $B=\left\{B_1,B_2,\dots ,B_{100}\right\}$.

- Student $B_1$ is friends with all students in $A$. - Student $B_2$ is friends with all students from $A_2$ to $A_{200}$. - ... - Student $B_{100}$ is friends with students from $A_{100}$ to $A_{200}$. - Students in the same group are not friends.

So it is easy to check that there are no any triplet of students that are friends with each other. And the number of friends of students in $A$ is $1,2,\dots ,100$ and the number of friends of students in $B$ is $200,199,\dots ,101$.