Let us denote with $S(n)$ the sum of all the digits of $n$. 
a) Is there such an $n$ that $n+S(n)=1980$? 
b) Prove that at least one of two arbitrary successive natural numbers is representable as $n + S(n)$ for some third number $n$.
1. We need to determine if there exists an integer \( n \) such that \( n + S(n) = 1980 \), where \( S(n) \) denotes the sum of the digits of \( n \).

2. First, observe that \( S(n) \equiv n \pmod{9} \). This is because the sum of the digits of a number is congruent to the number itself modulo 9. Therefore, we have:
   \[
   n + S(n) \equiv n + n \equiv 2n \pmod{9}
   \]
   Given \( n + S(n) = 1980 \), we can write:
   \[
   1980 \equiv 2n \pmod{9}
   \]
   Since \( 1980 \equiv 0 \pmod{9} \) (because the sum of the digits of 1980 is \( 1 + 9 + 8 + 0 = 18 \), and \( 18 \equiv 0 \pmod{9} \)), we have:
   \[
   2n \equiv 0 \pmod{9}
   \]
   This implies:
   \[
   n \equiv 0 \pmod{9}
   \]
   Therefore, \( n \) must be a multiple of 9.

3. Next, we need to find a multiple of 9 such that \( n + S(n) = 1980 \). Let \( n = 9k \) for some integer \( k \). Then:
   \[
   9k + S(9k) = 1980
   \]
   We need to find \( k \) such that this equation holds. We will test values of \( k \) to find a suitable \( n \).

4. Let's start with \( k = 218 \) (since \( 9 \times 218 = 1962 \)):
   \[
   n = 1962
   \]
   Calculate \( S(1962) \):
   \[
   S(1962) = 1 + 9 + 6 + 2 = 18
   \]
   Then:
   \[
   1962 + 18 = 1980
   \]
   This satisfies the equation \( n + S(n) = 1980 \).

Therefore, there exists an \( n \) such that \( n + S(n) = 1980 \), specifically \( n = 1962 \).

The final answer is \( \boxed{\text{yes, there is}} \)