If all the children have the same gender, then nobody can have a brother as well as a sister. In that case the number of children that match the condition is 0, regardless of $n$.

If there are children of either gender, but for at least one gender there is exactly one child of that gender, then this child does not have a brother (if he's a boy) or a sister (if she's a girl). If there is exactly one child of the other gender as well ($n = 2$), then the number of children that fulfill the condition is 0. If $n \ge 3$, then all other children have both a brother and a sister and there are $n-1$ children that satisfy the condition given in the problem statement. If there are at least 2 children of either gender, then all children have both a brother as well as a sister, and there are $n$ children satisfying the condition. This can happen when $n \ge 4$.

Summing up, for $n \le 2$ the answer is 0, for $n = 3$ there can be 0 or $n-1$ (in other words 2) such children and for $n \ge 4$ there are either 0, $n-1$ or $n$ such children.

For some child to have both a brother and a sister, there must be at least 3 children in this family. Therefore if $n \le 2$, then the answer is 0. Now assume that $n \ge 3$. It is clear that children of the same gender either all satisfy the condition given in problem statement or none of them do. In order for children of some gender not to satisfy the condition there must either be 1 child of such gender or 0 of the opposite gender. In the first case there are $n-1$ children satisfying the condition (since $n \ge 3$ and therefore there must be at least 2 children of the opposite gender), in the second case there are 0 such children. Therefore for $n \ge 3$ the possible answers are 0, $n-1$, and $n$. It suffices to note that children of both gender can only satisfy the condition if $n \ge 4$.