22nd Junior Turkish Mathematical Olympiad

The answer is $43890=1995\cdot 22$. Let us divide the students into two groups consisting of $1995$ and $22$ students. If each student played a match with each student of other group and no matches are played between students from the same group then the conditions are satisfied and there are $1995\cdot 22$ matches.

Let us show that total number of matches can not exceed $1995\cdot 22$. A student is said to be active if she played more than $22$ matches. Note that no match is played between two active students. If the number of non-active students is not exceeding $1995$ then the total number of matches is at most $1995\cdot 22$.

Now suppose that the number of non-active students is $1995+k$ for some positive integer $k$. Then the number of active students is $22-k$. The total number of matches is at most $(1995+k)\cdot 22$. The total number of matches involving active students is at most $(22-k)(1995+k)$. The remaining matches are played between non-active students and are counted twice. Therefore, the total number of matches is at most $$\frac{(1995+k)\cdot 22-(22-k)(1995+k)}{2}+(22-k)(1995+k)=\frac{(1995+k)\cdot 22+(22-k)(1995+k)}{2}=1995\cdot 22-\frac{k^2+1951k}{2}\le 1995\cdot 22.$$