Find the number of ways that a positive integer $n$ can be represented as a sum of one or more consecutive positive integers.
1. **Understanding the Problem:**
   We need to find the number of ways a positive integer \( n \) can be represented as a sum of one or more consecutive positive integers. 

2. **Formulating the Problem:**
   Let \( n \) be represented as a sum of \( k \) consecutive integers starting from \( a \). The sum of \( k \) consecutive integers starting from \( a \) is given by:
   \[
   a + (a+1) + (a+2) + \cdots + (a+k-1)
   \]
   This can be simplified using the formula for the sum of an arithmetic series:
   \[
   \sum_{i=0}^{k-1} (a+i) = ka + \frac{k(k-1)}{2}
   \]
   Therefore, we have:
   \[
   n = ka + \frac{k(k-1)}{2}
   \]

3. **Rewriting the Equation:**
   Rearrange the equation to solve for \( a \):
   \[
   n = ka + \frac{k(k-1)}{2} \implies 2n = 2ka + k^2 - k \implies 2n = k(2a + k - 1)
   \]
   Thus, we have:
   \[
   2n = k(2a + k - 1)
   \]

4. **Finding Integer Solutions:**
   For \( a \) to be an integer, \( 2a + k - 1 \) must be an integer. Let \( b = 2a + k - 1 \). Then:
   \[
   2n = kb \implies b = \frac{2n}{k}
   \]
   Here, \( b \) must be a positive integer, and \( k \) must be a divisor of \( 2n \).

5. **Counting the Solutions:**
   We need to count the number of pairs \((k, b)\) such that \( k \) is a divisor of \( 2n \) and \( b = \frac{2n}{k} \) is an integer. Additionally, \( b \) must be odd for \( a \) to be an integer:
   \[
   2a = b - k + 1 \implies a = \frac{b - k + 1}{2}
   \]
   For \( a \) to be a positive integer, \( b - k + 1 \) must be even, which means \( b \) must be odd.

6. **Conclusion:**
   The number of ways \( n \) can be represented as a sum of one or more consecutive positive integers is equal to the number of odd divisors of \( n \).

The final answer is the number of odd divisors of \( \boxed{ n } \).