Little couples in a school! Hmmm

Note: The answer is intuitive and it's fairly easy to check that it's the correct answer, but it's a little more complicated to provably find the answer. I've shown the deduction below, but note that we could have found the answer more easily.

Put $B_i$ boys and $G_i$ girls into the $i$ th classroom. We make the following two observations:

• If classes $i\neq j$ have $33 < B_i \leq G_i$ and $33 < G_j \leq B_j$ , then there are $B_i + G_j$ pairs in these two classes. However, by trading one boy from the $i$ th class with one girl from the $j$ th class, we would still have admissible classes, but with $B_i-1 + G_j-1 = B_i+G_j - 2$ pairs between these two classes.
• If classes $i\neq j$ have $33 < G_i < 60$ and $33 < G_j < 60$ with $G_i < G_j$ , then there are $G_i+G_j$ pairs in these two classes. Also, we can trade one girl from the $i$ th class with one boy from the $j$ th class to have $G_i - 1 \geq 33$ girls in the $i$ th class and $G_j+1 \leq 60$ girls in the $j$ th class, meaning there are still $G_i + G_j$ pairs between the two classes. Obviously, this argument also works for the boys in place of the girls.

Combining these, we can make the conclusion that the minimum occurs when at most one class $i$ has $B_i, G_i > 33$ . In other words, at least four classes have $33$ pairs. If all five classes had $33$ pairs, then we would have $k$ classes with $33$ boys and $5-k$ classes with $87$ boys giving $33k + 87(5-k) = 300 \implies k=\frac{5}{2},$ which is absurd since it isn't an integer. That is, exactly four classes have $33$ pairs.

Suppose $k$ of these four classes have $33$ girls and the last class has $G$ girls with $33<G<87$ . Then this gives $33k + 87(4-k) + G = 300 \implies 48+G = 54k \implies 48+33 < 54k < 48+87 \implies k=2 \text{ and } G = 54(2)-48 = 60.$ This means the fifth class has $120-60=60$ boys, and therefore $60$ pairs.

Finally, we have exhaustively shown that the minimum number of pairs that could be formed across all such schools (which is also the maximum we can guarantee can be formed) is $4(33) + 60 = \boxed{192}$

It is easy to find that each class has (300 + 300) ÷ 5 = 120 students.

In a worst case scenario (with the least number of pairs), there are 2 classes (A and B) having the minimum number of boys (33 boys and 120 - 33 = 87 girls; 33 pairs) each, another 2 classes (C and D) having the minimum number of girls (87 boys and 33 girls; 33 pairs) and the fifth class (E) having the rest of the students (60 boys and 60 girls, 60 pairs).

Now it is easy to see, that:

1.) The number of pairs only changes when we exchange a boy from one class for a girl from another class

2.) Such an exchange is not possible between the same type of classes ((A and B) or (C and D)) as it would result in less, than the mimimum number of boys/girls in one of the classes.

3.) A possible exchange between a class in the first group (A, B) and another class from the second group (C, D), e. g. a girl from A for a boy from D would increase the number of pairs in each class by one (by 2 in total)

4.) A possible exchange between the fifth class and another class (e.g. a boy from E for a girl from B, or a girl from E for a boy from C) would also result in the increase of the number of pairs in each class by 1 (2 in total)

This means, that we have now proven that our scenario results in the least number of possible pairs (which is the same as the maximum number of guaranteed pairs), and our answer should be:

$4 × 33 + 60 = \boxed {192}$