Consider the property that for integers $n \geq 3$, the divisors of $n!$, listed in increasing order as $1 = d_1 < d_2 < \dots < d_k = n!$, satisfy:

$$d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}.$$

To solve this problem, we analyze the differences $d_{i+1} - d_i$ for the sequence of divisors of $n!$.

### Step 1: Analysis for $n = 3$

Calculate $3! = 6$. The divisors of $6$ are $1, 2, 3, 6$.

- Differences: $d_2 - d_1 = 2 - 1 = 1$, $d_3 - d_2 = 3 - 2 = 1$, $d_4 - d_3 = 6 - 3 = 3$.

Check the condition:
$$1 \leq 1 \leq 3$$

The condition is satisfied for $n = 3$.

### Step 2: Analysis for $n = 4$

Calculate $4! = 24$. The divisors of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$.

- Differences:
$$\begin{align*} d_2 - d_1 & = 2 - 1 = 1, \\ d_3 - d_2 & = 3 - 2 = 1, \\ d_4 - d_3 & = 4 - 3 = 1, \\ d_5 - d_4 & = 6 - 4 = 2, \\ d_6 - d_5 & = 8 - 6 = 2, \\ d_7 - d_6 & = 12 - 8 = 4, \\ d_8 - d_7 & = 24 - 12 = 12. \end{align*}$$

Check the condition:
$$1 \leq 1 \leq 1 \leq 2 \leq 2 \leq 4 \leq 12$$

The condition is satisfied for $n = 4$.

### Step 3: Analysis for $n \geq 5$

For $n \geq 5$, consider the additional smaller prime divisors that appear in $n!$. These introduce smaller gaps among the divisors of $n!$, potentially violating the increasing condition of differences.

For example, for $n = 5$, $5! = 120$. The divisors include numbers like 10, 20, 30, etc., introducing nonuniform differences between consecutive divisors. This results in some differences being smaller than preceding differences, violating the original condition.

### Conclusion

The condition is satisfied only for $n = 3$ and $n = 4$, as detailed in the stepwise analysis. Therefore, the solution is:

$$\boxed{3 \text{ and } 4}$$