For any positive integer $n$, define a function $f$ by \[f(n)=2n+1-2^{\lfloor\log_2n\rfloor+1}.\] Let $f^m$ denote the function $f$ applied $m$ times.. Determine the number of integers $n$ between $1$ and $65535$ inclusive such that $f^n(n)=f^{2015}(2015).$

[i]Proposed by Yannick Yao[/i]

1. **Understanding the function \( f(n) \):**
   The function \( f(n) \) is defined as:
   \[
   f(n) = 2n + 1 - 2^{\lfloor \log_2 n \rfloor + 1}
   \]
   To understand this function, let's break it down:
   - \( \lfloor \log_2 n \rfloor \) gives the largest integer \( k \) such that \( 2^k \leq n \).
   - Therefore, \( 2^{\lfloor \log_2 n \rfloor + 1} \) is the smallest power of 2 greater than \( n \).

2. **Behavior of \( f(n) \):**
   - For a given \( n \), \( f(n) \) essentially subtracts the smallest power of 2 greater than \( n \) from \( 2n + 1 \).
   - This can be interpreted as a transformation in binary representation. Specifically, it affects the highest bit of \( n \).

3. **Binary Representation Insight:**
   - Consider the binary representation of \( n \). The function \( f(n) \) will eventually reduce \( n \) to a number with a sequence of 1's in binary.
   - For example, if \( n \) is 2015, its binary representation is \( 11111011111_2 \).

4. **Fixed Points of \( f \):**
   - A fixed point of \( f \) is a number that remains unchanged when \( f \) is applied. For \( f^m(n) \) to be equal to \( f^{2015}(2015) \), \( n \) must have the same number of 1's in its binary representation as 2015.

5. **Counting the Numbers:**
   - The binary representation of 2015 is \( 11111011111_2 \), which has 10 ones.
   - We need to count the number of integers between 1 and 65535 (which is \( 2^{16} - 1 \)) that have exactly 10 ones in their binary representation.

6. **Combinatorial Calculation:**
   - The number of ways to choose 10 positions out of 16 to place 1's (with the rest being 0's) is given by the binomial coefficient:
   \[
   \binom{16}{10}
   \]

7. **Final Calculation:**
   - Using the binomial coefficient formula:
   \[
   \binom{16}{10} = \frac{16!}{10!(16-10)!} = \frac{16!}{10! \cdot 6!}
   \]
   - Calculating this gives:
   \[
   \binom{16}{10} = 8008
   \]

The final answer is \(\boxed{8008}\).