Almost divisible by all 2

Let's first suppose that $n=k^2$ . Then it must be that $k^2 \equiv 0 \mod k-1$ , but $k^2 \equiv 1 \mod k-1$ so a perfect square never has this property for $k > 2$ . (It works for k=1 trivially and for k=2 because $1 \equiv 0 \mod 1$ ).

Let's now suppose that $k^2 < n < (k+1)^2$ . Then we must have that $k | n$ and $k-1 | n$ . Because $\gcd(k,k-1) = 1$ , it must be that $n$ is a multiple of $k(k-1)$ (assuming $k>1$ ). We can now consider the question of how many multiples of $k(k-1)$ can be between $k^2$ and $(k+1)^2$ . Let's consider two cases:

First $n = k(k-1)$ . This does not work because $k(k-1) < k^2$ .

Second, $n = 2k(k-1)$ . Then $2k(k-1) < (k+1)^2 \iff 2k^2 - 2k < k^2 + 2k + 1 \iff k^2 - 4k < 1 \iff (k-2)^2 < 5 \iff k-2 \leq 2 \iff k \leq 4$ . So for $k>4$ there are no solutions, and because we are looking the biggest solution, we must only check $k=4$ . Here $k(k-1) = 12, 2k(k-1) = 24, 3k(k-1) = 36$ . Only $24$ satisfies the inequalities, and we can quickly check that indeed $1,2,3,4 | 24$ .