Equally Divisible

Let $n$ be such a number (not necessarily the largest, but, being divisible by every number from $2$ to $\lfloor \sqrt{n} \rfloor$ .

Let $m$ be the largest integer less than or equal to $\sqrt{n}$ . Then $m^2 \le n < (m+1)^2$ .

Since, $n$ is a multiple of $m$ and should be within this range, we would have the choices of $n$ being restricted to $n=\{m^2,m(m+1),m(m+2)\}$ .

However, $n$ should also be divisible by $(m-1)$ . Hence, $n \equiv 0 \mod (m-1)$

Considering the three cases :

If $n=m^2$ , $m^2 \mod (m-1) = (m \mod (m-1))^2 = 1 \equiv 0 \mod (m-1)$ which means that $m-1=1 \implies m=2$ and $n=4$ .

If $n=m(m+1)$ , $m(m+1) \mod (m-1)=1 \times 2 = 2 \equiv 0 \mod (m-1)$ which means that $m-1=1,2 \implies m=2,3$ and $n=6,12$ .

If $n=m(m+2)$ , $m(m+2) \mod (m-1)=1 \times 3 = 3 \equiv 0 \mod (m-1)$ which means that $m-1=1,3 \implies m=2,4$ and $n=8,24$ .

These are the only choices for $n$ . Of these the maximum is $\boxed{24}$ .