Find all natural numbers $ n $ for which there exists two natural numbers $ a,b $ such that
$$ n=S(a)=S(b)=S(a+b) , $$
where $ S(k) $ denotes the sum of the digits of $ k $ in base $ 10, $ for any natural number $ k. $


[i]Vasile Zidaru[/i] and [i]Mircea Lascu[/i]
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 \( k \).