1. Claim: The answer is all integers $n$ of the form $\boxed{n=9k}$ for some positive integer $k$.

2. 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 $S(k) \equiv k \pmod{9}$. This is because the sum of the digits of $k$ in base 10 is congruent to $k$ modulo 9.

- Given $n = S(a) = S(b) = S(a+b)$, this implies:
$$a \equiv b \equiv a+b \pmod{9}$$
- From $a \equiv a+b \pmod{9}$, we get:
$$a \equiv 2a \pmod{9} \implies a \equiv 0 \pmod{9}$$
- This implies that $a$ is divisible by 9, and hence $S(a)$ is also divisible by 9. Therefore, $n$ must be divisible by 9.

3. Existence: Next, we show that if $n = 9k$ for some positive integer $k$, then $a$ and $b$ exist.

- 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, $n = S(a) = S(b) = S(a+b)$.

Therefore, the solution is all integers $n$ of the form $9k$ for some positive integer $k$.