Determine all non negative integers $k$ such that there is a function $f : \mathbb{N} \to \mathbb{N}$ that satisfies
\[ f^n(n) = n + k \]
for all $n \in \mathbb{N}$
1. **Base Case: \( k = 0 \)**
   - If \( k = 0 \), the equation \( f^n(n) = n + k \) simplifies to \( f^n(n) = n \).
   - This implies that \( f(n) = n \) for all \( n \in \mathbb{N} \).
   - Therefore, \( k = 0 \) is a valid solution.

2. **Assume \( k > 0 \)**
   - Suppose there exists a function \( f \) such that \( f^n(n) = n + k \) for some \( k > 0 \).
   - Consider the sequence \( A = \{a_h\}_{h \ge 0} = \{f^h(1)\}_{h \ge 0} \).

3. **No Cycles in \( f \)**
   - For any \( n, m \in \mathbb{N} \), \( f^m(n) \neq n \) because \( f^m(n) \) generates infinitely many distinct natural numbers of the form \( n + k, n + 2k, \ldots \).
   - Therefore, all elements in \( A \) must be distinct.

4. **Lemma: Distribution of Terms in \( A \)**
   - **Statement**: For any \( x \in \mathbb{N} \), there exist \( O(\sqrt{N}) \) terms in the first \( N \) terms of \( A \) that are congruent to \( x \pmod{k} \).
   - **Proof**:
     - If no such term exists, the lemma holds trivially.
     - Otherwise, let \( a_g \) be the smallest term congruent to \( x \pmod{k} \).
     - We observe that \( a_{g + n} = a_g + nk \) for all \( n \ge 0 \).
     - The number of desired terms is the number of non-negative integers \( n \) such that \( g + n \le N \), which is \( O(\sqrt{N}) \).

5. **Contradiction**
   - From the lemma, the first \( n \) terms of \( A \) have \( O(\sqrt{n}) \) values.
   - This implies that the sequence \( A \) cannot have \( n \) distinct values, which contradicts the assumption that all elements in \( A \) are distinct.

6. **Conclusion**
   - The assumption that \( k > 0 \) leads to a contradiction.
   - Therefore, the only possible value for \( k \) is \( 0 \).

The final answer is \( \boxed{ k = 0 } \).