Prove that there exist regular tournament of every odd order but there is no regular tournament of even order.
Yes, induction works. Suppose you have a regular tournament of order $n=2k+1$. The following procedure produces a tournament of order $n+2$.
- Add two new vertices $v$ and $v'$.
- For $k$ of the old vertices $w$, add edges $w \to v$ and $v' \to w$ so that $id(w)=od(w)$ still holds.
- For the remaining $k+1$ old vertices $w$, add edges $v \to w$ and $w \to v'$ so that $id(w)=od(w)$ still holds.
- Then $od(v)-id(v)=1$ and $od(v')-id(v')=-1$. Add the edge $v'\to v$ to finish.