Determine all functions $ f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy:

$ f(x\plus{}f(y))\equal{}f(x)\plus{}y$ for all $ x,y \in \mathbb{N}$.
1. **Initial Substitution:**
   Set \( x = 0 \) and \( y = 0 \) in the given functional equation:
   \[
   f(0 + f(0)) = f(0) + 0 \implies f(f(0)) = f(0).
   \]
   This implies that \( f(f(0)) = f(0) \).

2. **Second Substitution:**
   Set \( x = 0 \) and \( y = f(0) \):
   \[
   f(0 + f(f(0))) = f(0) + f(0) \implies f(f(f(0))) = 2f(0).
   \]
   From step 1, we know \( f(f(0)) = f(0) \), so:
   \[
   f(f(0)) = 2f(0) \implies f(0) = 0.
   \]

3. **Simplification:**
   With \( f(0) = 0 \), substitute \( x = 0 \) in the original equation:
   \[
   f(0 + f(y)) = f(0) + y \implies f(f(y)) = y.
   \]
   This shows that \( f \) is an involution, meaning \( f \) is its own inverse.

4. **Substitution and Simplification:**
   Substitute \( y = f(y) \) in the original equation:
   \[
   f(x + f(f(y))) = f(x) + y.
   \]
   Since \( f(f(y)) = y \), we have:
   \[
   f(x + y) = f(x) + y.
   \]

5. **Cauchy's Functional Equation:**
   The equation \( f(x + y) = f(x) + y \) resembles Cauchy's functional equation. However, since \( f: \mathbb{N} \rightarrow \mathbb{N} \), we need to find a specific solution that fits within the natural numbers.

6. **Verification:**
   Assume \( f(x) = x \). Substitute \( f(x) = x \) into the original equation:
   \[
   f(x + f(y)) = f(x + y) = x + y,
   \]
   and
   \[
   f(x) + y = x + y.
   \]
   Both sides are equal, confirming that \( f(x) = x \) is a solution.

7. **Uniqueness:**
   To show that \( f(x) = x \) is the only solution, assume there exists another function \( g \) such that \( g(x + g(y)) = g(x) + y \). By similar steps, we would derive that \( g \) must also satisfy \( g(f(y)) = y \) and \( g(x + y) = g(x) + y \), leading to \( g(x) = x \).

Thus, the only function that satisfies the given functional equation is \( f(x) = x \).

The final answer is \( \boxed{ f(x) = x } \).