Croatia_2018

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 $$a=\overline{18\dots 18},b=\overline{72\dots 72},a+b=\overline{90\dots 90},$$ where each of the numbers has exactly $2k$ digits, we obtain a pair $(a,b)$ which satisfies $S(a)=S(b)=S(a+b)=9k=n$.