Does there exists a function $f : \mathbb{N} \longrightarrow \mathbb{N}$
\begin{align*} f \left( m+ f(n) \right) = f(m) +f(n) + f(n+1) \end{align*}
for all $m,n \in \mathbb{N}$ ?
To determine whether there exists a function \( f : \mathbb{N} \longrightarrow \mathbb{N} \) such that 
\[ f(m + f(n)) = f(m) + f(n) + f(n+1) \]
for all \( m, n \in \mathbb{N} \), we will consider two cases: the non-injective case and the injective case.

1. **Non-injective Case:**
   - Assume \( f \) is not injective. This means there exist \( m > m' \) such that \( f(m) = f(m') \).
   - From the given functional equation, we have:
     \[ f(m + 1) = f(m' + 1). \]
   - By induction, we can show that:
     \[ f(x) = f(x + m - m') \quad \forall x \geq m'. \]
   - This implies that \( f \) is periodic with period \( m - m' \) for \( x \geq m' \). Hence, \( f \) is bounded for \( x \geq m' \).
   - Suppose \( f \) is not identically zero. Then there exists some \( n_0 \) such that \( f(n_0) > 0 \).
   - By induction, we have:
     \[ f(k \cdot f(n_0)) = f(0) + k \cdot (f(n_0) + f(n_0 + 1)) \geq k \cdot f(n_0) \geq k \rightarrow +\infty \text{ as } k \rightarrow +\infty. \]
   - This contradicts the boundedness of \( f \). Therefore, \( f \equiv 0 \).

2. **Injective Case:**
   - Assume \( f \) is injective. This means \( f(a) = f(b) \) implies \( a = b \).
   - Consider the functional equation:
     \[ f((n+2) + f(n)) = f(n + f(n+1)). \]
   - Since \( f \) is injective, we must have:
     \[ (n+2) + f(n) = n + f(n+1). \]
   - Simplifying, we get:
     \[ f(n+1) = f(n) + 2. \]
   - By induction, we can show that:
     \[ f(n) = f(0) + 2n. \]
   - Substitute \( f(n) = f(0) + 2n \) into the original equation:
     \[ f(m + f(n)) = f(m) + f(n) + f(n+1). \]
   - This becomes:
     \[ f(m + f(0) + 2n) = f(0) + 2m + f(0) + 2n + f(0) + 2n + 2. \]
   - Simplifying, we get:
     \[ f(0) + 2m + 2f(0) + 4n = f(0) + 2m + f(0) + 2n + f(0) + 2n + 2. \]
   - This implies:
     \[ 0 = 2, \]
     which is a contradiction. Therefore, \( f \equiv 0 \) is the only function.

The final answer is \( \boxed{ f \equiv 0 } \).