, 2023

The answer is $n-1$.

Indeed, the main claim is that $an \bmod 101$ is larger than $(a+1)n \bmod 101$ if and only if there is a number divisible by $101$ between $an$ and $(a+1)n$. Denote by $f(x)$ the remainder of $x$ when divided by $101$. Note that $f(an) \ne f((a+1)n)$, otherwise $101 \mid n$, a contradiction. Put $an = 101k + f(an)$ and $(a+1)n = 101k' + f((a+1)n)$. Thus
$$n = (a + 1)n - an = 101(k' - k) + f((a + 1)n) - f(an).$$
Since $-101 < f((a+1)n) - f(an) < 101$, it is easy to check that $k' - k = 0$ or $1$.

* If $f(an) < f((a+1)n)$ then $k = k'$ and between $an$ and $(a+1)n$, there does not exist any multiple of $101$.
* If $f(an) > f((a+1)n)$ then $k' = k + 1$ and between $an$ and $(a+1)n$, the number $101k'$ is a unique multiple of $101$.

So all we are left with is calculating the quantity of non-zero numbers divisible by $101$ between $0$ and $100 \cdot n$. This quantity is $\lfloor 100n/101 \rfloor = n - 1$.

$\boxed{n-1}$