SAUDI ARABIAN IMO Booklet 2023

The answer is $n-1$.

Indeed, the main claim is that $an\mathrm{mod}101$ is larger than $(a+1)n\mathrm{mod}101$ 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 $(a+1)n=101k^{\prime }+f((a+1)n)$. Thus $$n=(a+1)n-an=101(k^{\prime }-k)+f((a+1)n)-f(an).$$ Since $-101<f((a+1)n)-f(an)<101$, it is easy to check that $k^{\prime }-k=0$ or $1$.

If $f(an)<f((a+1)n)$ then $k=k^{\prime }$ and between $an$ and $(a+1)n$, there does not exist any multiple of $101$. If $f(an)>f((a+1)n)$ then $k^{\prime }=k+1$ and between $an$ and $(a+1)n$, the number $101k^{\prime }$ 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 $\lfloor 100n/101\rfloor =n-1$.

$\boxed{n-1}$