The answer is $65$. Suppose that there are $x$ girl and $y$ boy students in the school. Obviously $xy \ge 21x + 12y$ is a necessary condition for the existence of the tournament with given conditions. Let us show that this inequality is also a sufficient condition for the existence of the tournament. Let us consider $x \times y$ score table of the tournament. We put $1$ (alternatively $\emptyset$) in the intersection of $m$-th row and $n$-th column if $m$-th girl wins (alternatively loses) the match with $n$-th boy. Let us prove that there is a table in which each row contains at least $21$ entries $1$ and each column contains at least $12$ entries $\emptyset$.

We start with the table where each entry of the first $21$ columns is $1$ and each other entry is $\emptyset$ and step by step move to the table satisfying the conditions. Suppose that $k$-th column contains less than $12$ $\emptyset$'s. Since the total number of $\emptyset$'s is not less than $12y$ then some $l$-th column should contain more than $12$ $\emptyset$'s. In this case we choose a row such that the intersection of this row with $k$-th and $l$-th columns are $1$ and $\emptyset$, respectively and switch these two entries. After each move some column with $\emptyset$ shortage gains one $\emptyset$ and the total number of $1$ entries of each row does not change. Therefore after finite number of moves we will get the desired table. Done.

Now let us find the minimal value of $x + y$. The inequality $(x - 12)(y - 21) \ge 12 \cdot 21$ is equivalent to $xy \ge 21x + 12y$. By AM-GM inequality we get
$$(x-12) + (y-21) \ge 2\sqrt{(x-12)(y-21)} \ge 2\sqrt{12 \cdot 21} > 31.$$
Thus, $x + y \ge 12 + 21 + 32 = 65$. Any pair out of $(26,39)$, $(27,38)$, $(28,37)$, $(29,36)$, $(30,35)$ with sum $65$ satisfies the condition $xy \ge 21x + 12y$. Done.