Find all one-to-one mappings $f:\mathbb{N}\to\mathbb{N}$ such that for all positive integers $n$ the following relation holds:
\[ f(f(n)) \leq \frac {n+f(n)} 2 . \]
1. **Claim**: We claim that \( f(n) = n \) is the only solution. To prove this, we need to show that \( f(n) \leq n \) for all \( n \in \mathbb{N} \).

2. **Assumption**: Suppose, for the sake of contradiction, that there exists some \( n \in \mathbb{N} \) such that \( f(n) > n \).

3. **Induction**: We will use induction to show that \( f^k(n) > f^{k-1}(n) \) for all \( k \in \mathbb{N} \). 
   - Base case: For \( k = 1 \), we have \( f^1(n) = f(n) \) and \( f^0(n) = n \). By assumption, \( f(n) > n \).
   - Inductive step: Assume \( f^k(n) > f^{k-1}(n) \) for some \( k \). Then, \( f^{k+1}(n) = f(f^k(n)) \). By the given condition:
     \[
     f(f^k(n)) \leq \frac{f^k(n) + f(f^k(n))}{2}
     \]
     Since \( f \) is injective and \( f^k(n) > f^{k-1}(n) \), it follows that \( f(f^k(n)) > f(f^{k-1}(n)) \). Therefore, \( f^{k+1}(n) > f^k(n) \).

4. **Sequence Definition**: Let \( (a_k)_{k \geq 0} \) be a sequence defined by \( a_k = f^k(n) \) for all \( k \in \mathbb{N} \), with \( a_0 = n \). By the induction step, \( a_k \) is strictly increasing.

5. **Boundedness and Discreteness**: Since \( f \) maps \( \mathbb{N} \) to \( \mathbb{N} \), the sequence \( (a_k) \) is bounded and discrete. Therefore, there must exist \( k, l \in \mathbb{N} \) such that \( a_k = a_l \) with \( k \neq l \).

6. **Injectivity Contradiction**: By the injectivity of \( f \), if \( a_k = a_l \), then \( f^{k-l}(n) = n \). This implies that \( a_{k-l-1} = a_1 \), which contradicts the strict monotonicity of the sequence \( (a_k) \).

7. **Conclusion**: Therefore, our assumption that \( f(n) > n \) must be false. Hence, \( f(n) \leq n \) for all \( n \in \mathbb{N} \).

8. **Equality**: Since \( f \) is a one-to-one mapping, the only possibility left is \( f(n) = n \) for all \( n \in \mathbb{N} \).

\(\blacksquare\)

The final answer is \( \boxed{ f(n) = n } \) for all \( n \in \mathbb{N} \).