Let $n>1$ be a natural number and $x_k{}$ be the residue of $n^2$ modulo $\lfloor n^2/k\rfloor+1$ for all natural $k{}$. Compute the sum \[\bigg\lfloor\frac{x_2}{1}\bigg\rfloor+\bigg\lfloor\frac{x_3}{2}\bigg\rfloor+\cdots+\left\lfloor\frac{x_n}{n-1}\right\rfloor.\]
We claim that the answer is $(n-1)^2$.
First of all, note that $\frac{n^2}{\left\lfloor \frac{n^2}{k}\right\rfloor+1} \in (k-1,k)$ whenever $k \le n$. Indeed, this is equivalent to
\[\frac{n^2}{k}<\left\lfloor \frac{n^2}{k}\right\rfloor+1<\frac{n^2}{k-1}.\]Here, the LHS is trivial and the RHS follows since $\frac{n^2}{k-1}=\frac{n^2}{k}+\frac{n^2}{k(k-1)}>\frac{n^2}{k}+1$ if $k \le n$.
Hence we get $x_k=n^2-(k-1)\left(\left\lfloor \frac{n^2}{k}\right\rfloor+1\right)$. Hence
\[\frac{x_k}{k-1}=\frac{n^2}{k-1}-\left\lfloor \frac{n^2}{k}\right\rfloor-1\]and hence
\[\left\lfloor \frac{x_k}{k-1}\right\rfloor=\left\lfloor \frac{n^2}{k-1}\right\rfloor-\left\lfloor \frac{n^2}{k}\right\rfloor-1.\]But then the whole sum telescopes to $n^2-n-(n-1)=(n-1)^2$.