Solution. If there is a girl who does not have any friends, we can exclude her (the number of children will decrease, but the condition of the problem will remain).

Let us have $m$ boys and $d$ girls. Draw a segment between a boy and a girl if they are friends. Then $5m$ segments are drawn from the boys, and at least $\frac{d(d+1)}{2}$ segments are drawn from the girls. Indeed, let's order the girls by the number of acquaintances in ascending order; then the first girl is friends with at least one boy, the second with at least two, and so on. We also note that from this it follows that $m \geqslant d$ (the last girl is friends with at least $d$ boys, so there must be at least $d$ boys).

Then

$$
\frac{d(d+1)}{2} \geqslant 5 m \geqslant 5 d
$$

from which $d \geqslant 9$ and $m \geqslant 9$. An example with 9 boys and 9 girls is easy to construct: let the ninth girl know all the boys, the eighth - all except the first, the first - only the first boy, the seventh - except the second and third, the second - only the second and third, and so on.
6. Each boy is friends with 5 girls, while all girls are friends with a different number of boys. What is the minimum number of children that can be in this group?