Let $F$ be the set of all functions $f : N \to  N$ which satisfy $f(f(x))-2 f(x)+x = 0$ for all $x \in N$.
Determine the set $A =\{ f(1989) | f \in F\}$.
1. **Establishing the set \( A \supseteq \{k : k \ge 1989\} \):**
   - Consider the function \( f(x) = x + k \) for some integer \( k \ge 0 \).
   - Substitute \( f(x) = x + k \) into the given functional equation \( f(f(x)) - 2f(x) + x = 0 \):
     \[
     f(f(x)) = f(x + k) = (x + k) + k = x + 2k
     \]
     \[
     2f(x) = 2(x + k)
     \]
     \[
     f(f(x)) - 2f(x) + x = (x + 2k) - 2(x + k) + x = x + 2k - 2x - 2k + x = 0
     \]
   - This shows that \( f(x) = x + k \) satisfies the functional equation for any \( k \ge 0 \).
   - Therefore, \( f(1989) = 1989 + k \) for any \( k \ge 0 \), which implies \( A \supseteq \{k : k \ge 1989\} \).

2. **Showing \( A \subseteq \{k : k \ge 1989\} \):**
   - Assume there exists a function \( f \in F \) such that \( f(n) < n \) for some \( n \in \mathbb{N} \).
   - Let \( n_0 \) be the smallest positive integer such that \( f(n_0) < n_0 \).
   - Set \( f(n_0) = n_0 - k \) where \( k \) is a positive integer.
   - Since \( f(n_0) < n_0 \), we have \( k \ge 1 \) and \( k \le n_0 - 1 \).
   - Substitute \( n_0 \) into the functional equation:
     \[
     f(f(n_0)) - 2f(n_0) + n_0 = 0
     \]
     \[
     f(n_0 - k) - 2(n_0 - k) + n_0 = 0
     \]
     \[
     f(n_0 - k) = n_0 - 2k
     \]
   - Since \( n_0 \) is the smallest integer such that \( f(n_0) < n_0 \), \( f(n_0 - k) \ge n_0 - k \).
   - However, \( f(n_0 - k) = n_0 - 2k \) implies \( n_0 - 2k \ge n_0 - k \), which simplifies to \( -2k \ge -k \) or \( k \le 0 \), a contradiction since \( k \ge 1 \).
   - Therefore, \( f(n) \ge n \) for all \( n \in \mathbb{N} \).

3. **Conclusion:**
   - Since \( f(n) \ge n \) for all \( n \in \mathbb{N} \), and \( f(1989) = 1989 + k \) for some \( k \ge 0 \), we have \( A \subseteq \{k : k \ge 1989\} \).

Combining both inclusions, we conclude:
\[
A = \{k : k \ge 1989\}
\]

The final answer is \( \boxed{ \{k : k \ge 1989\} } \).