Find all functions $f: N_0 \to N_0$, (where $N_0$ is the set of all non-negative integers) such that $f(f(n))=f(n)+1$ for all $n \in N_0$ and the minimum of the set $\{ f(0), f(1), f(2) \cdots \}$ is $1$.
1. We start with the given functional equation:
   \[
   f(f(n)) = f(n) + 1 \quad \forall n \in \mathbb{N}_0
   \]
   and the condition that the minimum of the set \(\{ f(0), f(1), f(2), \ldots \}\) is 1.

2. Let's denote \(m\) as the minimum value of the set \(\{ f(0), f(1), f(2), \ldots \}\). According to the problem, \(m = 1\). Therefore, there exists some \(k \in \mathbb{N}_0\) such that \(f(k) = 1\).

3. Consider \(f(f(k))\). By the given functional equation:
   \[
   f(f(k)) = f(k) + 1 = 1 + 1 = 2
   \]
   Thus, \(f(1) = 2\).

4. Next, consider \(f(f(1))\):
   \[
   f(f(1)) = f(1) + 1 = 2 + 1 = 3
   \]
   Thus, \(f(2) = 3\).

5. We can generalize this pattern. Suppose \(f(n) = n + 1\) for some \(n \in \mathbb{N}_0\). Then:
   \[
   f(f(n)) = f(n + 1)
   \]
   By the functional equation:
   \[
   f(f(n)) = f(n) + 1 = (n + 1) + 1 = n + 2
   \]
   Therefore, \(f(n + 1) = n + 2\).

6. By induction, we can show that \(f(n) = n + 1\) for all \(n \in \mathbb{N}_0\):
   - Base case: For \(n = 0\), we have \(f(0) = 1\), which satisfies the condition.
   - Inductive step: Assume \(f(k) = k + 1\) for some \(k \in \mathbb{N}_0\). Then:
     \[
     f(k + 1) = (k + 1) + 1 = k + 2
     \]
     Thus, by induction, \(f(n) = n + 1\) for all \(n \in \mathbb{N}_0\).

7. Finally, we verify that \(f(n) = n + 1\) satisfies the original functional equation:
   \[
   f(f(n)) = f(n + 1) = (n + 1) + 1 = n + 2
   \]
   and
   \[
   f(n) + 1 = (n + 1) + 1 = n + 2
   \]
   Both sides are equal, confirming that \(f(n) = n + 1\) is a solution.

Conclusion:
\[
\boxed{f(n) = n + 1}
\]