Every positive integer can be represented as a sum of one or more consecutive positive integers. For each $n$ , find the number of such  represententation of $n$.
1. **Definitions and Initial Setup:**
   - Define a *Positive Sum* as a sum of consecutive positive integers that equals \( n \).
   - Define a *Negative Sum* as a sum of consecutive integers (including non-positive integers) that equals \( n \).

2. **Constructing Negative Sums from Positive Sums:**
   - Consider a Positive Sum \( n = a + (a+1) + (a+2) + \cdots + (a+k) \), where \( a \) is a whole number.
   - Construct a corresponding Negative Sum: \( n = -a + -(a-1) + -(a-2) + \cdots + 0 + 1 + 2 + \cdots + a + (a+1) + (a+2) + \cdots + (a+k) \).
   - Example: For the Positive Sum \( 2 + 3 + 4 \), the corresponding Negative Sum is \( -1 + 0 + 1 + 2 + 3 + 4 \).

3. **Constructing Positive Sums from Negative Sums:**
   - Reverse the process to construct a Positive Sum from a Negative Sum.
   - This shows that the number of Positive Sums is equal to the number of Negative Sums.

4. **Number of Sums:**
   - We claim that the number of sums (both Positive and Negative) is twice the number of divisors of the odd part of \( n \).
   - Let \( n = a + (a+1) + (a+2) + \cdots + (a+k) \). The sum can be written as:
     \[
     n = \frac{k}{2} \cdot (2a + k + 1)
     \]
   - Therefore, \( k \mid 2n \).

5. **Parity Considerations:**
   - From \( 2a + k + 1 = \frac{2n}{k} \), we get:
     \[
     2a = \frac{2n}{k} - k - 1
     \]
   - Since \( a \) is an integer, \( \frac{2n}{k} \) and \( k \) must have different parity.
   - At least one of \( \frac{2n}{k} \) or \( k \) is even.

6. **Divisors of \( n \):**
   - Assume \( n = 2^t (2l+1) \), where \( t \) and \( l \) are non-negative integers.
   - Then \( 2n = 2^{t+1} (2l+1) \).
   - If \( k \) is odd, \( k \mid 2l+1 \).
   - If \( k \) is even, \( 2^{t+1} \mid k \) and \( k \mid 2^{t+1}(2l+1) \).

7. **Counting Divisors:**
   - The number of even \( k \) equals the number of odd \( k \).
   - The number of odd \( k \) is \( \tau(2l+1) \), where \( \tau \) is the number of divisors function.
   - Therefore, the total number of sums is \( 2 \cdot \tau(2l+1) \).

8. **Conclusion:**
   - The number of Positive Sums is \( \tau(2l+1) \), the number of divisors of the odd part of \( n \).

The final answer is \( \boxed{ \tau(2l+1) } \).