Problem: Find all integers $n \geq 3$ such that the following property holds:
if we list the divisors of $n!$ in increasing order
as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]

Solution: The answer is $n \in \{3,4\}$. These can be checked by listing all the divisors:
\begin{itemize}
  \ii For $n=3$ we have $(1,2,3,6)$.
  \ii For $n=4$ we have $(1,2,3,4,6,8,12,24)$.
\end{itemize}

We prove these are the only ones.

The numbers $5 \le n \le 12$ all fail because:
\begin{itemize}
  \ii For $n = 5$ we have $20 - 15 > 24 - 20$.
  \ii For $n = 6$ we have $120 - 90 > 144 - 20$.
  \ii For $7 \le n \le 12$ we have because $14-12 > 15-14$ (and $13 \nmid n!$).
\end{itemize}

Now assume $n \ge 13$.
In that case, we have
\[ \left\lfloor \frac n2 \right\rfloor^2 - 1 \ge 2n. \]
So by Bertrand postulate, we can find a prime $p$ such that
\[ n < p < \left\lfloor \frac n2 \right\rfloor^2 - 1. \]
However, note that
\[ \left\lfloor \frac n2 \right\rfloor^2 - 1
  = \left( \left\lfloor \frac n2 \right\rfloor - 1 \right) \left( \left\lfloor \frac n2 \right\rfloor + 1 \right),
  \qquad \left\lfloor \frac n2 \right\rfloor^2
\]
are consecutive integers both dividing $n!$ (the latter number divides
$\left\lfloor \frac n2 \right\rfloor \cdot \left( 2 \left\lfloor \frac n2 \right\rfloor \right)$).
So the prime $p$ causes the desired contradiction.