Hermite's identity for floor functions

1. Put $x=N+f$ with $N=\lfloor x\rfloor$ and $0\le f<1$. Then $\lfloor x+\frac kn\rfloor=N+\lfloor f+\frac kn\rfloor$, and $\lfloor f+\frac kn\rfloor\in\{0,1\}$. It equals $1$ iff $f+\frac kn\ge1$, i.e. $k\ge n-nf$, equivalently $k\ge n-\lfloor nf\rfloor$.

So among $k=0,\dots,n-1$, the number of terms equal to $N+1$ is $\lfloor nf\rfloor$. Hence the left side is
$$nN+\lfloor nf\rfloor=\lfloor nN+nf\rfloor=\lfloor nx\rfloor.$$

2. Take $x=\frac mn$: then $\lfloor nx\rfloor=m$, and the identity becomes $\sum_{k=0}^{n-1}\left\lfloor\frac{m+k}n\right\rfloor=m$.

Example. $m=100$, $n=7$: the terms are $14,14,14,14,14,15,15$, whose sum is $100$ ✓.

Lesson. To handle floor functions, split into integer and fractional parts and find where the fractional part crosses an integer threshold.