Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that \[f(n+1)>\frac{f(n)+f(f(n))}{2}\] for all $n\in\mathbb{N}$, where $\mathbb{N}$ is the set of strictly positive integers.
1. **Initial Assumption and Trivial Case**:
   - We start by noting that \( f(n) \ge n \) for all \( n \in \mathbb{N} \). This is because if \( f(n) < n \), then \( f(n+1) > \frac{f(n) + f(f(n))}{2} \) would not hold as \( f(f(n)) \le f(n) < n \), leading to a contradiction.
   - We also observe that \( f(n) = n \) is a solution. Let's assume \( f(n) \neq n \) for some \( n \).

2. **Defining the Difference**:
   - Consider the sequence \( f(1) - 1, f(2) - 2, \ldots, f(n) - n \). Let \( k \) be the maximum value of these differences.
   - Let \( b \) be the smallest number such that \( f(b) - b = k \).

3. **Finding a Contradiction**:
   - Suppose there exists an \( a > b \) such that \( f(a+1) - (a+1) < k \).
   - We have:
     \[
     f(a+1) > \frac{f(a) + f(f(a))}{2}
     \]
   - Substituting \( f(a) = a + k \) and \( f(f(a)) = f(a + k) \), we get:
     \[
     f(a+1) > \frac{a + k + f(a + k)}{2}
     \]
   - Since \( f(a + k) \ge a + k \), we have:
     \[
     f(a+1) > \frac{a + k + (a + k)}{2} = a + k
     \]
   - This implies \( f(a+1) > a + k \), which contradicts the assumption that \( f(a+1) - (a+1) < k \).

4. **Conclusion for \( f(n) = n + k \)**:
   - Therefore, after some number \( b \), every number will satisfy \( f(n) = n + k \).
   - Consider \( n > b \). We have:
     \[
     f(n+1) > \frac{f(n) + f(f(n))}{2}
     \]
   - Substituting \( f(n) = n + k \) and \( f(f(n)) = f(n + k) = n + k + k \), we get:
     \[
     n + k + 1 > \frac{n + k + n + 2k}{2} = n + \frac{3k}{2}
     \]
   - Simplifying, we get:
     \[
     n + k + 1 > n + \frac{3k}{2}
     \]
     \[
     k + 1 > \frac{3k}{2}
     \]
     \[
     2(k + 1) > 3k
     \]
     \[
     2k + 2 > 3k
     \]
     \[
     2 > k
     \]
   - Thus, \( k < 2 \), so \( k \) can only be 1 (since \( k \) is a non-negative integer).

5. **Final Form of the Function**:
   - Therefore, the function will be \( f(n) = n \) for \( n < b \) and \( f(n) = n + 1 \) for \( n \ge b \).

The final answer is \( \boxed{ f(n) = n } \) for \( n < b \) and \( f(n) = n + 1 \) for \( n \ge b \).