Find all positive integers $n$ such that $n$ is divisible by all the positive integers less than or equal to $\sqrt{n}$ Find all positive integers $n$ such that $n$ is divisible by all the
positive integers less than or equal to $\sqrt{n}$
My thought:
If n is a positive integer, let d(n) denote the number of positive
divisors of n and I will find n such that $d(n) \ge 2\sqrt{n} - 1$

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 $$1, 2, 3, 4, 6, 8, 12, 24.$$
