When does the first repetition in $\;\lfloor x\rfloor, \lfloor x/2 \rfloor, \lfloor x/3\rfloor, \lfloor x/4\rfloor, \dots\;$ appear? Let $\lfloor x\rfloor$ denote the floor of $x$.

When does the first repetition in $\lfloor x\rfloor$, $\lfloor x/2\rfloor$, $\lfloor x/3\rfloor$, $\lfloor x/4\rfloor$, ... approximately appear, as a function of $x$?

It seems to be around ~ $c \sqrt x$.
Example: $x = 2500$:
2500, 1250, 833, 625, 500, 416, 357, 312, 277, 250, 227, 208, 192, 178, 166, 156, 147, 138, 131, 125, 119, 113, 108, 104, 100, 96, 92, 89, 86, 83, 80, 78, 75, 73, 71, 69, 67, 65, 64, 62, 60, 59, 58, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 43, 42, 41, 40, 40, 39, 39, 38, 37, 37, 36, 36, 35, 35, ...

$\color{brown}{\textbf{Preliminary Notes.}}$
$\dfrac xN > \dfrac x{N+1},$ so the required equality 
$$\left\lfloor\dfrac xN\right\rfloor = \left\lfloor\dfrac x{N+1}\right\rfloor,\tag1$$
where $\lfloor a\rfloor = \mathrm{floor}\,(a),$
has solutions iff
$$\left\{\dfrac xN\right\} > \left\{\dfrac x{N+1}\right\}.\tag2$$
Taking in account that
$$k\left\{\dfrac xk\right\} = x\hspace{-12mu}\mod k,$$
inequality $(2)$ can be presented in the form of
$$\dfrac{x\hspace{-12mu}\mod N}N > \dfrac{x\hspace{-12mu}\mod (N+1)}{N+1},\tag3$$
and from $(3)$ should
$$x\hspace{-12mu}\mod N \ge x\hspace{-12mu}\mod (N+1).\tag4$$
Formula $(4)$ can be used for the testing, instead the issue one.
$\color{brown}{\textbf{Decision.}}$
The least solution $N$ of equality $(1)$ belongs to the interval $x\in(n(n-1),n(n+1)],$ so 
$$x=n(n-1)+m,\quad m=x-(n^2-n),\quad m\in[1,2n],\tag5$$
wherein
$$n = \left\lceil\dfrac{\sqrt{4x-3}+1}2\right\rceil\tag6,$$
$\lceil a\rceil = \mathrm{ceil}\,(a).$
Let
$$N=n+k,\quad k\in\mathbb N\tag7,$$
$$\dfrac xN = \dfrac{n^2-n+m}{n+k}=n-k-1+\dfrac{k(k+1)+m}{n+k},$$
$$\dfrac x{N+1} = \dfrac{n^2-n+m}{n+k+1}=n-k-2+\dfrac{(k+1)(k+2)+m}{n+k+1},$$
If $\underline{m\in[1,n-1]}$ then the least solution of $(3)$ can be achieved iff
$$(k+1)(k+2)+m\ge n+k+1,\quad k =\left\lceil\sqrt{n-m}-1\right\rceil,$$
$$N = n - 1 + \left\lceil\sqrt{n^2-x}\right\rceil.$$
If $\underline{m\in[n,2n-1]}$ then the least solution of $(3)$ can be achieved iff
$$(k+1)(k+2)+m\ge 2(n+k+1),\quad k^2+k-(2n-m)\geq 0,\quad k = \left\lceil\dfrac{\sqrt{8n-4m+1}-1}2\right\rceil,$$
$$N = n + \left\lceil\dfrac{\sqrt{4(n^2+n-x^2)+1}-1}2\right\rceil.$$
If $\underline{m=2n}$ then 
$$\dfrac xN = \dfrac{n^2+n}{n+k}=n-k+1+\dfrac{k(k-1)}{n+k},$$
$$\dfrac x{N+1} = \dfrac{n^2+n}{n+k+1}=n-k+\dfrac{k(k+1)}{n+k+1},$$
and the least solution of $(3)$ can be achieved iff
$$k(k-1)<n+k,\quad (k-1)^2<n+1,\quad k = \left\lceil\sqrt{n+1}\right\rceil,$$
$$N = n + \left\lceil\sqrt{n+1}\right\rceil.$$
$\color{brown}{\textbf{Result.}}$
Therefore, the least solution of $(1)$ is
$$\boxed{\begin{align}
&N=\begin{cases}
n - 1 +\left\lceil\sqrt{n^2-x}\right\rceil,\quad\text{if}\quad x-n(n-1)\in[1,n-1]\\[4pt]
n + \left\lceil\dfrac{\sqrt{4(n^2+n-x)+1}-1}2\right\rceil,\quad\text{if}\quad x-n(n-1)\in[n,2n-1]\\[4pt]
n + \left\lceil\sqrt{n+1}\right\rceil,\quad\text{if}\quad x = n(n+1),
\end{cases}\\
&\text{where}\\
&n = \left\lceil\dfrac{\sqrt{4x-3}+1}2\right\rceil.
\end{align}}\tag8$$
$\color{brown}{\textbf{Examples.}}$
$\underline{x=2475,\quad n=50,\quad x-n(n-1)=25}.$
There is the first case of $(8).$
Result is $N=54,$ with $\left\lfloor\dfrac{2475}{54}\right\rfloor = \left\lfloor\dfrac{2475}{55}\right\rfloor=45.$
$\underline{x=2500,\quad n=50,\quad x-n(n-1)=50}.$
There is the second case of $(8).$
Result is $N=57,$ with $\left\lfloor\dfrac{2500}{57}\right\rfloor = \left\lfloor\dfrac{2500}{58}\right\rfloor=43.$
$\underline{x=2450,\quad n=49,\quad x=n(n+1)}.$
There is the third case of $(8).$
Result is $N=57,$ with $\left\lfloor\dfrac{2450}{57}\right\rfloor = \left\lfloor\dfrac{2450}{58}\right\rfloor=42.$
