My solution suffices for $r\ge 0$. Denote $\{x\}$ to be the fractional part of $x$, in other words, $x=\lfloor x\rfloor+\{x\}$. First, I hope its obvious that if $r\in\mathbb{Z},$ then the theorem holds. We now proceed by contradiction. If $r$ is not an integer, then $r$ can be written in the form $z+\epsilon$ where $z\in\mathbb{Z}$ and $0<\epsilon<1$. Now, using the bounds on $\epsilon$, we have that there exists an integral $m\ge 3$ such that $2<m\epsilon<3$. Then $\lfloor mr\rfloor=\lfloor (z+\epsilon)m\rfloor=zm+\lfloor m\epsilon\rfloor=mz+2\ge 2$. Now, lets consider the number $\{m\epsilon\}$. Since $2<m\epsilon<3,$ we know that $0<\{m\epsilon\}<1$. Thus, there exists an integral $k\ge 2$ such that $1\le k\{m\epsilon\}<2$. Let $n=mk$. Then: $\displaystyle\lfloor nr\rfloor=\lfloor (km)(z+\epsilon)\rfloor=kmz+\lfloor km\epsilon\rfloor=kmz+\lfloor k\lfloor m\epsilon\rfloor + k\{m\epsilon\}\rfloor=k(mz+\lfloor m\epsilon\rfloor) +1$ But then $\displaystyle\frac{k(mz+\lfloor m\epsilon\rfloor)+1}{mz+\lfloor m\epsilon\rfloor}=k+\frac{1}{mz+\lfloor m\epsilon\rfloor}$ and since the denominator is greater than or equal to $2,$ the ratio cannot be an integer. Therefore $\epsilon=0$ and $r\in\mathbb{Z}$.

Let $ m,k$ be positive integers and $ r\ge 0$ with the mentioned property. So there exists a integer $ l$ such that $ \lfloor mkr\rfloor=l\lfloor mr\rfloor$ Let $ p=\lfloor mr\rfloor$ and $ q=mr-p$, then $ kp+\lfloor kq\rfloor =lp\Rightarrow p(l-k)=\lfloor kq\rfloor$. If $ p=0$ we must have $ q=0$ so $ r$ is an integer. Suppose $ p>0$. Since $ 0\le q<1$ we have $ 0\le \frac{\lfloor kq\rfloor}{p}\le\frac{k-1}{p}$. This must hold for all $ m$ and $ k$ so let's take $ k=p$, it becomes $ 0\le \frac{\lfloor kq\rfloor}{p}=l-k\le\frac{p-1}{p}<1$ then $ q=0$, again $ r$ is an integer. This completes the proof.

Let $k$ be an integer and $r$ with the desired property. Pick $m=1$ and write $\lfloor kr \rfloor = \lfloor k \lfloor r \rfloor + k \langle r \rangle \rfloor $, where $\langle r \rangle$ is the decimal part of $r$, notice that since $k \lfloor r \rfloor$ is and integer, we can take it out from the floor function. Rearranging the terms we get: \[ \lfloor kr \rfloor - \lfloor k \langle r \rangle \rfloor = k \lfloor r \rfloor \]. If $\lfloor k \langle r \rangle \rfloor=1,$ then $\lfloor kr \rfloor$ is not a multiple of $\lfloor r \rfloor $. But we can always find $k$ that satisfies this if $\langle r \rangle >0$, just by taking the least number such that $\langle r \rangle>1$ we instantly see that $2>k \langle r \rangle>1$ and so, no matter what $r$ we take, if it has a non-zero decimal there exists a $k$ for which the divisibility condition doesn't hold, thus $r$ is an integer.

Let $r\ge 1$ be a real number that holds with the property that for each pair of positive integer numbers $m$ and $n$, with $n$ a multiple of $m$, it is true that $\lfloor nr \rfloor$ is multiple of $\lfloor mr \rfloor$. Show that $r$ has to be an integer number. Note: If $x$ is a real number, $\lfloor x \rfloor$ is the greatest integer lower than or equal to $x$.

Assume $\lfloor {mr} \rfloor \nmid \lfloor nr \rfloor$ and $r \not\in \mathbb{Z}$. Then $\exists$ unique $k \in \mathbb{Z}$ such that \begin{align*} \displaystyle \frac{1}{k+1}\leq mr-\lfloor mr \rfloor<\frac{1}{k} \displaystyle \implies &1\leq (mr-\lfloor mr \rfloor)(k+1) < \frac{k+1}{k}\leq 2\\\implies &\lfloor(mr-\lfloor mr \rfloor)(k+1) \rfloor=1 \\\Longleftrightarrow &\lfloor(k+1)mr\rfloor-1=(\lfloor mr \rfloor)(k+1) \rfloor \end{align*}On dividing both sides by $\lfloor mr \rfloor$ we get that $\lfloor mr \rfloor \mid 1 \implies \lfloor mr \rfloor =1$ which is a contradiction to our prior assumption because $1 \mid \lfloor nr \rfloor$.