Find all injective functions $ f : \mathbb{N} \to \mathbb{N}$ which satisfy
\[ f(f(n)) \le\frac{n \plus{} f(n)}{2}\]
for each $ n \in \mathbb{N}$.
To find all injective functions \( f : \mathbb{N} \to \mathbb{N} \) that satisfy the inequality
\[ f(f(n)) \le \frac{n + f(n)}{2} \]
for each \( n \in \mathbb{N} \), we will proceed as follows:

1. **Consider the identity function**:
   Let \( f(n) = n \). Then,
   \[ f(f(n)) = f(n) = n \]
   and
   \[ \frac{n + f(n)}{2} = \frac{n + n}{2} = n. \]
   Therefore,
   \[ f(f(n)) = n \le n, \]
   which satisfies the given inequality. Hence, the identity function \( f(n) = n \) is a solution.

2. **Assume \( f \) is not the identity function**:
   If \( f \) is not the identity function, there exists some \( n \in \mathbb{N} \) such that \( f(n) \neq n \). We consider two cases: \( f(n) > n \) and \( f(n) < n \).

3. **Case 1: \( f(n) > n \)**:
   If \( f(n) > n \), then
   \[ f(f(n)) \le \frac{n + f(n)}{2} < f(n). \]
   This implies that \( f(f(n)) < f(n) \). Since \( f \) is injective, \( f(n) \) must be distinct for different \( n \), and thus \( f(f(n)) \) must be less than \( f(n) \).

4. **Case 2: \( f(n) < n \)**:
   If \( f(n) < n \), then
   \[ f(f(n)) \le \frac{n + f(n)}{2} < n. \]
   This implies that \( f(f(n)) < n \). By induction, if \( f^k(n) < n \) for any \( 1 \le k \le m \), then
   \[ f^{m+1}(n) \le \frac{f^m(n) + f^{m-1}(n)}{2} < n. \]
   This means all iterations of \( f \) applied to \( n \) will be less than \( n \).

5. **Cycle formation**:
   Since \( n \) is finite, the iterations will either form a cycle or stop at some point. Suppose they form a cycle of the form \( a_1, a_2, \ldots, a_t \) where \( f(a_i) = a_{i+1} \) and \( f(a_t) = a_1 \). Consider the greatest \( a_i \), say \( a_M \). Then,
   \[ a_{M+1} = f(a_M) \le \frac{a_M + a_{M-1}}{2}. \]
   Since \( a_{M-1} \neq a_M \) and \( a_{M-1} < a_M \), it follows that \( a_{M+1} < a_M \). Repeating this process, we find that all iterations of \( f(a_i) \) for \( i \neq M \) will be strictly less than \( a_M \), and therefore will not return to \( a_M \).

6. **Contradiction**:
   If the iterations stop at some point, let the last term be \( a_m \). Then \( f(a_{m-1}) = a_m \) and \( f(a_m) = a_m \). Since \( f \) is injective, \( a_{m-1} \neq a_m \) is impossible. Therefore, \( f(n) = n \) for all \( n \).

Conclusion:
The only injective function \( f : \mathbb{N} \to \mathbb{N} \) that satisfies the given inequality is the identity function.

\(\blacksquare\)

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