To solve the problem, we need to find the maximum number of subsets of $\{1, 2, \ldots, n\}$ such that for any two subsets $A$ and $B$, if $A \subset B$, then $|B - A| \geq 3$. Here, $|X|$ denotes the number of elements in the set $X$.

1. Initial Observation:
- Consider the set $\{1, 2, \ldots, n\}$.
- We need to ensure that for any two subsets $A$ and $B$ where $A \subset B$, the difference in their sizes is at least 3.

2. Grouping Subsets:
- Let's group subsets based on their sizes. Specifically, consider subsets with sizes $3i+1$ (if $n$ is even) or $3i+2$ (if $n$ is odd).
- This grouping ensures that the difference in sizes between any two subsets in the group is at least 3.

3. Counting Subsets:
- The number of subsets of size $3i+1$ or $3i+2$ can be approximated by summing the binomial coefficients $\binom{n}{3i+1}$ or $\binom{n}{3i+2}$.
- The sum of these binomial coefficients is given by:
$$\sum_{i=0}^{\left\lfloor \frac{n}{3} \right\rfloor} \binom{n}{3i+1} \quad \text{or} \quad \sum_{i=0}^{\left\lfloor \frac{n}{3} \right\rfloor} \binom{n}{3i+2}$$
- This sum is approximately $\frac{2^n - (-1)^n}{3}$.

4. Inductive Proof:
- Base Case: For $n = 4$, we can manually verify that the maximum number of subsets satisfying the condition is 6. The subsets are $\{1, 2\}$, $\{2, 3\}$, $\{1, 3\}$, $\{1, 4\}$, $\{2, 4\}$, and $\{3, 4\}$.
- Inductive Step: Assume the result holds for $n-1$. We need to prove it for $n$.
- Split the subsets into two groups: those that do not contain the element 1 and those that do.
- The first group has at most $\frac{2^{n-1} - (-1)^{n-1}}{3}$ subsets by the inductive hypothesis.
- The second group can be considered by removing the element 1 from each subset, resulting in at most $\frac{2^{n-1} - (-1)^{n-1}}{3} + 1$ subsets (including the subset $\{1\}$).
- Combining these, we get the total number of subsets as $\frac{2^{n-1} - (-1)^{n-1}}{3} + \frac{2^{n-1} - (-1)^{n-1}}{3} + 1 = \frac{2^n - (-1)^n}{3}$.

5. Correction:
- For $n = 4$, the actual number of subsets satisfying the condition is 6, not 5.
- Therefore, the correct maximum number of subsets is $\left\lceil \frac{2^n}{3} \right\rceil$.

The final answer is $\left\lceil \frac{2^n}{3} \right\rceil$.