Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n$ ! in increasing order as $1=d_{1}<d_{2}<\cdots<d_{k}=n$ !, then we have  
$$ d_{2}-d_{1} \leq d_{3}-d_{2} \leq \cdots \leq d_{k}-d_{k-1} $$
The answer is $n \in\{3,4\}$. These can be checked by listing all the divisors:  
- For $n=3$ we have $(1,2,3,6)$.  
- For $n=4$ we have $(1,2,3,4,6,8,12,24)$.  
We prove these are the only ones. The numbers $5 \leq n \leq 12$ all fail because:  
- For $n=5$ we have $20-15>24-20$.  
- For $n=6$ we have $18-15>20-18$.  
- For $7 \leq n \leq 12$ we have because $14-12>15-14$ (and $13 \nmid n!$ ).  
Now assume $n \geq 13$. In that case, we have  
$$ \left\lfloor\frac{n}{2}\right\rfloor^{2}-1 \geq 2 n $$  
So by Bertrand postulate, we can find a prime $p$ such that  
$$ n<p<\left\lfloor\frac{n}{2}\right\rfloor^{2}-1 $$  
However, note that  
$$ \left\lfloor\frac{n}{2}\right\rfloor^{2}-1=\left(\left\lfloor\frac{n}{2}\right\rfloor-1\right)\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right), \quad\left\lfloor\frac{n}{2}\right\rfloor^{2} $$  
are consecutive integers both dividing $n$ ! (the latter number divides $\left\lfloor\frac{n}{2}\right\rfloor \cdot\left(2\left\lfloor\frac{n}{2}\right\rfloor\right)$ ). So the prime $p$ causes the desired contradiction.