Find all $a\in\mathbb{N}$ such that exists a bijective function $g :\mathbb{N} \to \mathbb{N}$ and a function $f:\mathbb{N}\to\mathbb{N}$, such that for all $x\in\mathbb{N}$, 
$$f(f(f(...f(x)))...)=g(x)+a$$ where $f$ appears $2009$ times.

[i](tatari/nightmare)[/i]
1. **Observation on Injectivity and Range**:
   - We need to find a function \( f \) such that \( f^{2009} \) (the function \( f \) composed with itself 2009 times) is injective and its range is precisely the naturals larger than \( a \).
   - The injectivity of \( f^{2009} \) implies the injectivity of \( f \).

2. **Structure of \( f \)**:
   - The digraph of \( f \) over \( \mathbb{N} \), due to injectivity, is a union of some finite cycles and some infinite chains.
   - Let there be \( k \) chains \( C_i \) and let \( C_{ij} \) be the \( j \)-th element of the \( i \)-th chain.

3. **Elements Not in the Image of \( f^{2009} \)**:
   - The numbers that are not in the image of \( f^{2009} \) are precisely the numbers \( C_{ij} \) for \( 1 \leq i \leq k \) and \( 1 \leq j \leq 2009 \).
   - All elements of cycles are images of \( f^{2009} \) anyway.

4. **Equation for \( a \)**:
   - Thus, the number of elements not in the image of \( f^{2009} \) is \( 2009k \).
   - This forces \( 2009 \mid a \).

5. **Existence of Functions for Multiples of 2009**:
   - For each \( a \) that is a multiple of 2009, such functions exist.
   - We can set \( f(n) = n + \frac{a}{2009} \) and \( g(n) = n \) for all \( n \in \mathbb{N} \).

6. **Verification**:
   - For \( f(n) = n + \frac{a}{2009} \), we have:
     \[
     f^{2009}(n) = n + 2009 \cdot \frac{a}{2009} = n + a
     \]
   - This satisfies the given condition \( f(f(f(...f(x)))...) = g(x) + a \) where \( f \) appears 2009 times and \( g(x) = x \).

Conclusion:
\[
\boxed{a \text{ must be a multiple of } 2009}
\]