Let $A = \{ a_1, a_2, a_3, a_4 \}$ be a set of four distinct positive integers. We define $s_A = a_1 + a_2 + a_3 + a_4$ as the sum of these integers. We also define $n_A$ as the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ such that $a_i + a_j$ divides $s_A$.

Our goal is to find all sets $A$ for which $n_A$ is maximized. Notice that there are a total of ${4 \choose 2} = 6$ pairs $(i, j)$. The maximum possible value of $n_A$ is 6, which occurs when all pairs divide $s_A$.

To maximize $n_A$, consider the sums $a_i + a_j$ for all pairs, and ensure that these sums divide the total sum $s_A$. A useful observation is that if all elements are multiples of a common factor, some division properties become more regular. Thus, consider elements in arithmetic form based on a common ratio $k$.

Check a possible set $\{ k, 5k, 7k, 11k \}$:
1. Compute $s_A = k + 5k + 7k + 11k = 24k$.
2. Consider possible values of $i, j$ and verify divisibility:
- $a_1 + a_2 = k + 5k = 6k$, $24k \div 6k = 4$.
- $a_1 + a_3 = k + 7k = 8k$, $24k \div 8k = 3$.
- $a_1 + a_4 = k + 11k = 12k$, $24k \div 12k = 2$.
- $a_2 + a_3 = 5k + 7k = 12k$, $24k \div 12k = 2$.
- $a_2 + a_4 = 5k + 11k = 16k$, $24k \div 16k = 1.5$; not integer.
- $a_3 + a_4 = 7k + 11k = 18k$, $24k \div 18k$ is not an integer multiple.

Realize that the pair $(a_2 + a_4)$ doesn't divide $s_A$. Now consider a permuted set: $\{ k, 11k, 19k, 29k \}$:
1. Compute $s_A = k + 11k + 19k + 29k = 60k$.
2. Analyze pairs:
- $a_1 + a_2 = k + 11k = 12k$, $60k \div 12k = 5$.
- $a_1 + a_3 = k + 19k = 20k$, $60k \div 20k = 3$.
- $a_1 + a_4 = k + 29k = 30k$, $60k \div 30k = 2$.
- $a_2 + a_3 = 11k + 19k = 30k$, $60k \div 30k = 2$.
- $a_2 + a_4 = 11k + 29k = 40k$, $60k \div 40k = 1.5$; again not an integer.
- $a_3 + a_4 = 19k + 29k = 48k$, $60k \div 48k$ not an integer multiple.

Given new attempts, both sets achieve maximum $n_A = 5$, which is the highest possible given constraints.

Therefore, the sets $\{ k, 5k, 7k, 11k \}$ and $\{ k, 11k, 19k, 29k \}$ are valid solutions that achieve the largest possible $n_A$.

Final answer:
$$\boxed{\{k, 5k, 7k, 11k\} \quad \text{and} \quad \{k, 11k, 19k, 29k\}}$$