We first observe how the height of people changes as we go around the circle in the clockwise direction. For every pair of neighbours $A$ and $B$ (where $B$ is after $A$ in the clockwise direction) we put the symbol ♣ between them if $B$ is taller than $A$ and the symbol ♠ if $B$ is shorter than $A$. In that way we obtain a sequence of $N$ symbols ♣ or ♠.
A person is average if and only if the symbol before that person is equal to the symbol after that person. Hence the number of people who are not average is equal to the number of alternations of the symbols ♣ and ♠. The number of changes from the symbol ♣ to ♠ is equal to the number of changes from ♠ to ♣. So the number of people who are not average is even.
From this we conclude that the number of average people is of the same parity as $N$. Since the tallest person in the circle is not average the number of average people is certainly smaller than $N$.
Let $a_1 < a_2 < \dots < a_N$ denote the heights of people. We will exhibit an example showing that for all $k = 1, \dots, \lfloor N/2 \rfloor$ there exists a configuration of $N$ people in a circle such that exactly $N - 2k$ is average. More precisely, for a configuration where the people are ordered in a circle so that their heights are respectively
$$\underbrace{a_1, a_N, a_2, a_{N-1}, \dots, a_k, a_{N-k+1}}_{\text{A}}, \underbrace{a_{N-k}, \dots, a_{k+1}}_{\text{G}}$$
the first $2k$ people are not average and the remaining $N - 2k$ are average.
This shows that the number of average people can be any number strictly less than $N$ and having the same parity as $N$.