I claim that the answer is all integers $n$ of the form $\boxed{n=9k}$ for some positive integer $k$. Proof: First, we show that $n$ must be divisible by $9$ in order for the property to hold. Observe that for any positive integer $k$, we have $S(k)\equiv k\pmod9$. Therefore, since $$n=S(a)=S(b)=S(a+b)$$this implies that $$a\equiv b\equiv a+b\pmod9\implies a\equiv2a\pmod9\implies a\equiv0\pmod9$$This implies that $S(a)$ is divisible by $9$, and so is $n$. Next, we show that if $n=9k$ for some positive integer $k$, then $a$ and $b$ exists. Indeed, consider $a=b=\underbrace{909090\cdots909090}$, where the number has exactly $k$ copies of $90$. This makes $a+b=181818\cdots1818180$, where $a+b$ has exactly $k$ copies of $18$. This gives us $$S(a)=S(b)=S(909090\cdots909090)=(9+0)\cdot k=9k$$and $$S(a+b)=S(181818\cdots1818180)=(1+8)\cdot k+0=9k$$Hence, $n=S(a)=S(b)=S(a+b)$.

I claim that the answer is all multiples of 9. It is easy to see that $n=9k$ works when $a=b=90909090...909$, where there are $k 9’s$ Now, we show that $9|n$ Note that $S(a)=S(a+b)$ implies that $S(a) = S(a+b) \mod 9$ But then this implies that $a = S(a) = S(a+b) = a+b \mod 9$ Thus, $9|b$, so $9|S(b)=n$ Edit: sniped