Fix integer $d>1$, and assume real number $x\in[0,1]$.
I claim the following statement:
$\sum_{k=1}^{dn}\lfloor kx\rfloor=d\sum_{k=1}^n\lfloor kx\rfloor$ is true iff $x\in[0,\frac{1}{dn}]$.
I can check it in various cases, and the if-direction is obvious. But my combinatorics skills are not sharp, so is there a counterexample?
You must have $x \in \left[0,\frac{1}{dn}\right)$, for if $x = \frac{1}{dn}$, then the left hand side is $1$, and the right hand side is $0$.
Generally, if $\frac{1}{dn} \leqslant x < \frac{1}{n}$, then
$$d\sum_{k=1}^n \lfloor kx\rfloor = 0 < \lfloor dnx\rfloor \leqslant \sum_{k=1}^{dn} \lfloor kx\rfloor,$$
so in that case you don't have equality. And if $x \geqslant \frac1n$, then you have
$$\begin{align} \sum_{k=1}^{dn} \lfloor kx\rfloor - d\sum_{k=1}^n \lfloor k\rfloor &= \sum_{j=0}^{d-1} \left(\sum_{k=1}^n \left(\lfloor (k+jn)x\rfloor - \lfloor kx\rfloor\right)\right)\\ &\geqslant\sum_{j=0}^{d-1} \sum_{k=1}^n\left(\lfloor kx + j\rfloor - \lfloor kx\rfloor\right)\\ &\geqslant \sum_{j=0}^{d-1} nj\\ &= n\frac{d(d-1)}{2}\\ &> 0. \end{align}$$