Thank you for your help!!
Here's the question:
Find all positive integers n divisible by all positive integers m where m is less than or equal to n^(1/2)
First note that the only $\displaystyle n<36$ which fill the condition are
$\displaystyle n = 1, 2, 3, 4, 6, 8, 12, 24$. We'll see there are no others.
Suppose $\displaystyle n$ fills the conditions, and let $\displaystyle t = n-\lfloor \sqrt n \rfloor^2$.
We can suppose now that $\displaystyle \lfloor \sqrt n \rfloor \ge 6$ and then arrive at a contradiction.
$\displaystyle \lfloor \sqrt n \rfloor \mid n$, so there exists a positive integer $\displaystyle s$ such that
$\displaystyle s\lfloor \sqrt n \rfloor = n = \lfloor \sqrt n \rfloor^2+t$
Therefore $\displaystyle \lfloor \sqrt n \rfloor\mid t$ and we can write $\displaystyle n-\lfloor \sqrt n \rfloor^2 = t = d\lfloor \sqrt n \rfloor$
That is, $\displaystyle n = \lfloor \sqrt n \rfloor^2+d\lfloor \sqrt n \rfloor$.
Write $\displaystyle \lfloor \sqrt n \rfloor = \sqrt{n}-\epsilon,\:\: 0\le\epsilon<1$. Plugging in the previous equality yields:
$\displaystyle d = \epsilon\frac{2\sqrt{n}-\epsilon}{\sqrt{n}-\epsilon} < \frac{2\sqrt{n}-\epsilon}{\sqrt{n}-\epsilon} = 1+\frac{\sqrt{n}}{\sqrt{n}-\epsilon} < 1+ \frac{\sqrt{n}}{\sqrt{n}-2}$
And since $\displaystyle \sqrt{n}\ge 6$ we have $\displaystyle \frac{\sqrt{n}}{\sqrt{n}-2}<2$, so $\displaystyle d<1+2 = 3 \Rightarrow d\in\{0,1,2\}$
Thus $\displaystyle n\le h^2+2h$ where we denote $\displaystyle h=\lfloor \sqrt n \rfloor$
We have $\displaystyle h\mid n$ and $\displaystyle h-1\mid n$ and these numbers
are coprime so $\displaystyle h(h-1)\mid n$. Furthermore, $\displaystyle h-2\mid n$
and is coprime to $\displaystyle h-1$, and $\displaystyle \gcd(h-2,h)\le2$,
so $\displaystyle h(h-1)(h-2) \mid 2n$, and thus $\displaystyle h(h-1)(h-2) \le 2n \le 2h^2+4h$
Therefore $\displaystyle h^2-3h+2 \le 2h+4 \Rightarrow h^2-5h -2 \le 0 \Rightarrow h < 6$,
contradicting our assumption that $\displaystyle h=\lfloor\sqrt{n}\rfloor \ge 6 $ (see head of the post).