Leaving aside the trivial case $\{1\}$, a maximal special set may have only 3, 4 or 5 elements. Begin by noticing that if $a < b$ are positive odd integers, and $a$ does not divide $b$, then $a$, $b$ and $2b-a$ form a special set, so a maximal special set has at least three elements.

At the other extreme, a special set — in particular, one that is maximal — has at most five elements. The proof relies on the three facts below:

(1) If $a > b > c$ form a special set, then $b+c$ is not divisible by $a$. This is because $a$ is odd, and $b+c$ is a positive even integer less than $2a$.

(2) If $a > b$ are members of a special set $S$, then at most one of the members of $S$ less than $b$ does not divide the sum $a+b$. Suppose, if possible, $c$ and $d$ are distinct members of $S$ less than $b$, neither of which divides the sum $a+b$. By (1), $a$ divides neither $b+c$ nor $b+d$, so $a+c$ and $a+d$ are both divisible by $b$. Then $|c-d| = |(a+c) - (a+d)|$ is a positive integer less than $b$ and divisible by $b$ — a contradiction.

(3) If $a, b, c, d$ form a special set, and $a + b$ and $a + c$ are both divisible by $d$, then $b+c$ is not divisible by $d$. Otherwise, $d$ would divide $(a+b)+(a+c)-(b+c) = 2a$, which is impossible, since $d$ is odd and does not divide $a$.

We are now in a position to prove that a special set has at most five elements. Suppose, if possible, $a_1, a_2, a_3, b_1, b_2, b_3$ are pairwise distinct members of a special set. We may and will assume $a_1 > a_2 > a_3 > \max(b_1, b_2, b_3)$.

Fix a pair of distinct indices $i$ and $j$, and write $\{i, j, k\} = \{1, 2, 3\}$. By (2), some $b$ divides both $a_i + a_k$ and $a_j + a_k$, by (3), that $b$ does not divide $a_i + a_j$, so, with reference again to (2), it is the unique $b$ not dividing $a_i + a_j$.

Consequently, the three $b$'s may be labeled so that $b_i$ and $b_j$ both divide $a_i + a_j$, while $b_k$ does not, $\{i, j, k\} = \{1, 2, 3\}$.

By (1), $a_1$ does not divide $a_3 + b_2$, and since $b_2$ does not divide $a_1 + a_3$, it follows that $a_1 + b_2$ is divisible by $a_3$. Similarly, $a_2 + b_1$ is divisible by $a_3$, and hence so is $(a_1 + b_2) + (a_2 + b_1) = (a_1 + a_2) + (b_1 + b_2)$. Finally, since $a_1 + a_2$ is divisible by $a_3$, by (2), so is $b_1 + b_2$, in contradiction with (1).

Consequently, a special set — in particular, one that is maximal — has at most five elements.

Next, we show that 5-element special sets actually exist. Clearly, the numbers $3$, $5$, $7$ form a special set. To enlarge this set to a 4-element special set by adjoining a positive odd integer $k$, notice that $k$ divides no 2-term sum from $\{3, 5, 7\}$, to infer that $k$ satisfies one of the two systems of linear congruences below:

$$\begin{cases} k + 1 \equiv 0 \pmod{2} \\ k + 5 \equiv 0 \pmod{3} \\ k + 3 \equiv 0 \pmod{7} \\ k + 7 \equiv 0 \pmod{5} \end{cases} \quad \text{or} \quad \begin{cases} k + 1 \equiv 0 \pmod{2} \\ k + 3 \equiv 0 \pmod{5} \\ k + 5 \equiv 0 \pmod{7} \\ k + 7 \equiv 0 \pmod{3} \end{cases}$$

By the Chinese remainder theorem, each of these systems has infinitely many solutions; in each case, two solutions differ by a multiple of $2 \cdot 3 \cdot 5 \cdot 7 = 210$. The least positive solution of the former is $193$, and the least positive solution of the latter is $107$.

To enlarge the set $\{3, 5, 7, 193\}$ to a 5-element special set by adjoining a positive odd integer $k$, notice again that $k$ divides no 2-term sum from $\{3, 5, 7, 193\}$, to infer that $k$ satisfies the system of linear congruences

$$\begin{cases} k+1 \equiv 0 \pmod{2} \\ k+3 \equiv 0 \pmod{5} \\ k+5 \equiv 0 \pmod{7} \\ k+7 \equiv 0 \pmod{3} \\ k+7 \equiv 0 \pmod{193} \end{cases} ;$$

clearly, $3$ and $5$ both divide $k+193$. As before, the Chinese remainder theorem settles the case; incidentally, the least positive solution is $3467$, and all five numbers are prime.

Similarly, the set $\{3, 5, 7, 107\}$ extends to a 5-element special set by adjoining any positive odd integer $k$ satisfying the system of linear congruences

$$\begin{cases} k+1 \equiv 0 \pmod{2} \\ k+5 \equiv 0 \pmod{3} \\ k+3 \equiv 0 \pmod{7} \\ k+7 \equiv 0 \pmod{5} \\ k+7 \equiv 0 \pmod{107} \end{cases} ;$$

clearly, $3$ and $5$ both divide $k+107$. In this case, the least positive solution is $10693 = 17^2 \cdot 37$.

Remark. The 5-element special sets below are obtained in the same way:
$$\{3, 5, 13, 127, 17267 = 31 \cdot 557\}, \quad \{3, 5, 17, 97, 14353 = 31 \cdot 463\},$$
$$\{3, 7, 11, 235 = 5 \cdot 47, 26309\}, \quad \{3, 7, 11, 437 = 19 \cdot 23, 60295 = 5 \cdot 31 \cdot 389\}.$$

We now show that the special set consisting of $3$, $5$, $13$, $17$ is maximal. Suppose, if possible, that $k$ is a positive odd integer such that $3$, $5$, $13$, $17$, $k$ form a special set. It is easily seen that $k$ divides no 2-term sum from $\{3, 5, 13, 17\}$.

We first show that $k+5$ is divisible by $3$ if and only if $k+3$ is divisible by $5$; since one holds, so does the other.

If $k+5$ is divisible by $3$, then $k+13$ is not, so $k+3$ is divisible by $13$. It follows that $k+5$ is not divisible by $13$, so $k+13$ is divisible by $5$, showing that $k+3$ is indeed divisible by $5$.