Solve in Python:
Let's say we have a number, `num`. Find the number of values of `n` such that: there exists `n` consecutive **positive** values that sum up to `num`. A positive number is `> 0`. `n` can also be 1.

```python
#Examples
num = 1
#1
return 1

num = 15
#15, (7, 8), (4, 5, 6), (1, 2, 3, 4, 5)
return 4

num = 48
#48, (15, 16, 17)
return 2

num = 97
#97, (48, 49)
return 2
```
The upper limit is `$10^8$`
def consecutive_sum(num):
    count = 0
    N = 1
    while( N * (N + 1) < 2 * num): 
        a = (num - (N * (N + 1) ) / 2) / (N + 1) 
        if (a - int(a) == 0.0): 
            count += 1
        N += 1
    return count+1