For any positive integer $n$, let $a_n=\sum_{k=1}^{\infty}[\frac{n+2^{k-1}}{2^k}]$, where $[x]$ is the largest integer that is equal or less than $x$. Determine the value of $a_{2015}$.
1. **Understanding the Expression**:
   We need to evaluate the sum \( a_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n + 2^{k-1}}{2^k} \right\rfloor \) for \( n = 2015 \). Here, \( \left\lfloor x \right\rfloor \) denotes the floor function, which gives the largest integer less than or equal to \( x \).

2. **Analyzing the Summand**:
   Consider the term \( \left\lfloor \frac{n + 2^{k-1}}{2^k} \right\rfloor \). For large \( k \), specifically when \( 2^k > n + 2^{k-1} \), the term \( \frac{n + 2^{k-1}}{2^k} \) becomes less than 1, and thus its floor value is 0. Therefore, the sum effectively has a finite number of non-zero terms.

3. **Simplifying the Expression**:
   Let's rewrite the expression for clarity:
   \[
   \left\lfloor \frac{n + 2^{k-1}}{2^k} \right\rfloor = \left\lfloor \frac{n}{2^k} + \frac{2^{k-1}}{2^k} \right\rfloor = \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor
   \]
   Since \( \frac{1}{2} \) is less than 1, adding it to \( \frac{n}{2^k} \) does not change the integer part unless \( \frac{n}{2^k} \) is an integer.

4. **Counting the Terms**:
   For each \( k \), \( \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor \) is essentially counting how many times \( 2^{k-1} \) fits into \( n \) when considering the binary representation of \( n \). This is equivalent to summing the binary digits of \( n \).

5. **Summing the Series**:
   The sum \( \sum_{k=1}^{\infty} \left\lfloor \frac{n + 2^{k-1}}{2^k} \right\rfloor \) effectively counts each integer from 1 to \( n \) exactly once. This is because each integer \( m \) from 1 to \( n \) will be counted in the term where \( k \) is such that \( 2^{k-1} \leq m < 2^k \).

6. **Conclusion**:
   Therefore, the sum \( a_n \) is simply \( n \), as each integer from 1 to \( n \) is counted exactly once.

\[
a_{2015} = 2015
\]

The final answer is \(\boxed{2015}\)