Find all positive integers $n$ such that there exist $n$ consecutive positive integers whose sum is a perfect square.

1. **Sum of Consecutive Integers**:
   Consider \( n \) consecutive positive integers starting from \( k+1 \) to \( k+n \). The sum \( S \) of these integers can be expressed as:
   \[
   S = (k+1) + (k+2) + \cdots + (k+n)
   \]
   This is an arithmetic series with the first term \( k+1 \) and the last term \( k+n \). The sum of an arithmetic series is given by:
   \[
   S = \frac{n}{2} \left( (k+1) + (k+n) \right)
   \]
   Simplifying the expression inside the parentheses:
   \[
   S = \frac{n}{2} \left( 2k + n + 1 \right)
   \]
   Therefore:
   \[
   S = nk + \frac{n(n+1)}{2}
   \]

2. **Case 1: \( n \) is Odd**:
   Let \( n = 2m-1 \) where \( m \) is a positive integer. We need to find \( k \) such that \( S \) is a perfect square. Consider:
   \[
   S = n \left( k + \frac{n+1}{2} \right)
   \]
   Substituting \( n = 2m-1 \):
   \[
   S = (2m-1) \left( k + \frac{2m-1+1}{2} \right) = (2m-1) \left( k + m \right)
   \]
   To make \( S \) a perfect square, set \( k = (2m-1)s^2 - m \) for some integer \( s \):
   \[
   S = (2m-1) \left( (2m-1)s^2 - m + m \right) = (2m-1)^2 s^2
   \]
   Thus, \( S \) is a perfect square.

3. **Case 2: \( n \) is Even**:
   Let \( n = 2^u \cdot v \) where \( v \) is odd and \( u > 0 \). We need to find \( k \) such that \( S \) is a perfect square. Consider:
   \[
   S = n \left( k + \frac{n+1}{2} \right)
   \]
   Substituting \( n = 2^u \cdot v \):
   \[
   S = 2^u \cdot v \left( k + \frac{2^u \cdot v + 1}{2} \right)
   \]
   Simplifying:
   \[
   S = 2^{u-1} \cdot v \left( 2k + v \cdot (2^u \cdot v + 1) \right)
   \]
   For \( S \) to be a perfect square, \( v_2(S) = u-1 \) must be even. Hence, \( u \) must be odd. Set \( k = \frac{(2l+1)^2 - v \cdot (2^u \cdot v + 1) - 1}{2} \) for some integer \( l \):
   \[
   S = 2^{\frac{u-1}{2}} \cdot l
   \]
   Thus, \( S \) is a perfect square.

4. **Conclusion**:
   The only \( n \) for which there exist \( n \) consecutive positive integers whose sum is a perfect square are those for which \( v_2(n) \) is either \( 0 \) or an odd number.

The final answer is \( \boxed{ n } \) such that \( v_2(n) \) is either \( 0 \) or an odd number.