I think this question can be solved by a high school student, maybe there is some trick on it or I'm forgetting something.
Before my question, some background is required:
Definition: A numerical semigroup is a subsemigroup $H$ of the additive semigroup $\mathbb N$ of the non-negative integers such that $\mathbb N-H$ is finite and $0\in H$. A numerical semigroup has genus $g$ if $L=\mathbb N-H$ has $g$ elements. The elements of $L$, $1=l_1\lt l_2\lt \ldots\lt l_g$, are called gaps of $H$, while the elements of $H$, $0=n_0\lt n_1\lt \ldots$ are called non-gaps of $H$. If $n_1=2$ the semigroup is called hyperelliptic.
Using this definition, I'm trying to understand this little piece of the proof of this theorem:
So, my question is why $l_j-(2j-r)\in H$? I almost sure it's a silly thing.
I realized after some days that perhaps this problem is genuinely a combinatorial one.
Thanks in advance
The reason why this is so is because of the statement in the previous paragraph that "if the equality holds then each pair $(n, \ell_j-n)$ has a gap and a non-gap" (I'm renaming the variable here to prevent confusion). Since we have that $k = 2j-r\in L$, it must be the case that $\ell_j-k\in H$.
Meanwhile, the reason for this statement is a basic pigeonhole argument: there are $j-1$ pairs of the form $(n, \ell_j-n)$ and only $j-1$ gaps less than $\ell_j$. Because no pair can have zero gaps (because then $n\in H$ and $\ell_j-n\in H$, so $\ell_j = n+(\ell_j-n)\in H$, contradiction), no pair can have more than one gap (or the total count of gaps would be $\gt j-1$); therefore, each pair has a gap and a non-gap.