First, you’re right in thinking that it’s essentially the same idea. It’s also basically correct, but it could stand a bit of reorganization. Specifically, it would be better to begin by applying the division algorithm to $\lfloor x\rfloor$ and $d$. And if I were giving a proof more detailed than the one in the book, I’d probably also go into a bit more detail about the manipulation of the floor function. Making those changes but otherwise trying to stay as close as possible to your basic approach, I might end up with something like this:
The positive integers less than or equal to $x$ are precisely the integers $1,2,\ldots,\lfloor x\rfloor$. By the division algorithm there are unique integers $q$ and $r$ such that $\lfloor x\rfloor=dq+r$, and $0\le r<d$, so we can split these $\lfloor x\rfloor$ positive integers into $q$ blocks of $d$ integers each, possibly with a short block of $r$ integers left over: $$\begin{align*}&1,2,\ldots,d,\\&d+1,d+2,\ldots,2d,\\
&2d+1,2d+2,\ldots,3d,\\&\qquad\qquad\vdots\\&(q-1)d+1,(q-1)d+2,\ldots,qd,\\&qd+1,\ldots,qd+r\end{align*}$$ The multiples of $d$ here are the integers at the ends of the first $q$ lines, namely, $d,2d,3d,\ldots,qd$. Clearly there are $q$ of them, so we need to show that $q=\left\lfloor\frac{x}d\right\rfloor$.
Let $\alpha=x-\lfloor x\rfloor$, the fractional part of $x$; of course $0\le\alpha<1$. Then $0\le r+\alpha<d$, so $$\left\lfloor\frac{x}d\right\rfloor=\left\lfloor\frac{dq+r+\alpha}d\right\rfloor=\left\lfloor q+\frac{r+\alpha}d\right\rfloor=q+\left\lfloor\frac{r+\alpha}d\right\rfloor=q\;,$$ as desired.
Best Answer
By manual testing, $n=1$, $n=2$, $n=4$, $n=6$, $n=8$ have the desired property and are the only numbers $<9$ with it.
Let $n$ be a natural number that is divisible by all natural numbers $\le \sqrt n$. So if we let $a=\lfloor \sqrt n\rfloor$ then $a$ and $a-1$ (if $>0$) are by divisors of $n$. If we additionally assume $n\ge 9$ then $a\ge 3$. Since $a$ and $a-1$ are relatively prime, $a^2-a=a(a-1)$ divides $n$. Thus $n=k(a^2-a)$ for some $k$. Clearly $k>1$ because $n\ge a^2$. Hence $(a+1)^2>n\ge 2a^2-2a$, i.e., $0\ge a^2-4a=(a-4)a$. This implies $a\le 4$. If $a=3$ we find $9\le n<16$ and $n$ must be a multiple of $2$ and $3$. We need only check $n=12$, which gives a solution. If on the other hand $a=4$, then $16\le n<25$ and $n$ is a multiple of $3$ and $4$. We need only check $n=24$, which gives a solution.
Thus the positive integers with the property are precisely $$1, 2, 3, 4, 6, 8, 12, 24.$$