Notice that, for each positive integer $a$, the numbers $a$ and $S(a)$ give the same remainder when divided by $9$.
Applying this observation to $S(a) = S(b) = S(a+b)$, we conclude that $a$, $b$ and $a+b$ give the same remainder when divided by $9$. This implies that $(a+b) - b = a$ and $(a+b) - a = b$ are both divisible by $9$. But then $a$ and $b$ are also divisible by $9$. This proves that any positive integer $n$ satisfying the given condition must be divisible by $9$.

Let us now prove the converse: let $n$ be a multiple of $9$, i.e. let $n = 9k$ for some positive integer $k$. Taking numbers
$$a = \overline{18\ldots18}, \quad b = \overline{72\ldots72}, \quad a+b = \overline{90\ldots90},$$
where each of the numbers has exactly $2k$ digits, we obtain a pair $(a, b)$ which satisfies $S(a) = S(b) = S(a+b) = 9k = n$.