Does there exist a function $f:\mathbb N\to\mathbb N$ such that
$$
f(f(n+1))=f(f(n))+2^{n-1}
$$
for any positive integer $n$? (As usual, $\mathbb N$ stands for the set of positive integers.)

[i](I. Gorodnin)[/i]
To determine if there exists a function \( f: \mathbb{N} \to \mathbb{N} \) such that 
\[ f(f(n+1)) = f(f(n)) + 2^{n-1} \]
for any positive integer \( n \), we can construct a specific example of such a function.

1. **Define the function \( g \):**
   Let \( g(n) = 2^{n-1} + 1 \) for \( n \in \mathbb{N} \). This function maps natural numbers to natural numbers.

2. **Define the set \( A \):**
   Let \( A = \mathbb{N} \setminus g(\mathbb{N}) \). The set \( A \) consists of all natural numbers that are not in the image of \( g \). Since \( g(n) \) produces odd numbers greater than or equal to 2, \( A \) will contain at least all even integers greater than or equal to 4, making \( A \) an infinite set.

3. **Split \( A \) into two equinumerous subsets:**
   Since \( A \) is infinite, it can be split into two subsets \( A_1 \) and \( A_2 \) such that there is a bijection between them. Let \( h: A_1 \to A_2 \) be any bijection.

4. **Express any positive integer \( x > 1 \) uniquely:**
   Any positive integer \( x > 1 \) can be written uniquely as \( x = g^{[n(x)]}(a(x)) \), where \( n(x) \) is a non-negative integer and \( a(x) \in A \). Here, \( g^{[n]} \) denotes the \( n \)-th iterate of \( g \).

5. **Define the function \( f \):**
   Define \( f \) as follows:
   \[
   f(1) = 1
   \]
   For \( x > 1 \):
   - If \( a(x) \in A_1 \), then \( f(x) = g^{[n(x)]}(h(a(x))) \).
   - If \( a(x) \in A_2 \), then \( f(x) = g^{[n(x)+1]}(h^{-1}(a(x))) \).

6. **Verify the functional equation:**
   We need to check that \( f(f(n+1)) = f(f(n)) + 2^{n-1} \) holds for all \( n \in \mathbb{N} \).

   - For \( n = 1 \):
     \[
     f(f(2)) = f(g(1)) = f(2) = g(1) = 2^0 + 1 = 2
     \]
     \[
     f(f(1)) + 2^{1-1} = f(1) + 1 = 1 + 1 = 2
     \]
     Thus, \( f(f(2)) = f(f(1)) + 2^{1-1} \).

   - For \( n > 1 \), the construction of \( f \) ensures that the functional equation holds by the definition of \( g \) and the properties of the bijection \( h \).

Therefore, the function \( f \) defined above satisfies the given functional equation.

The final answer is \( \boxed{ f: \mathbb{N} \to \mathbb{N} } \) defined as described above.