Claim: Any positive integer \(n\) can be written as \(n=2^k\cdot m\) , where \(k\ge0\) and \(m\) is an odd positive integer. To prove this claim use the fact :  \(n=2^k\cdot p_1^{k_1}\cdot p_2^{k_2}\cdots p_i^{k_i}\), where \(k\) and all \(k_i\) are non-negetive integers and all \(p_i\) are odd primes.

The sum of (any) \(r\) consecutive positive integers is given by, \((q+1)+(q+2)+(q+3)+\cdots+(q+r)\) =   \(q\cdot r+(1+2+3+\cdots+r)\) =   \(q\cdot r+\frac{r(r+1)}{2}\).  

Equating this sum to \(n\) we get,           \(2^k\cdot m = q\cdot r+\frac{r(r+1)}{2}\)   Or,     \(2^{k+1}\cdot m = 2q\cdot r+r(r+1)\)   Or,     \(2^l\cdot m = r(2q+r+1)\) , where \(l=k+1\ge1\).

In particular if we take \(n=2^l\) then \(m\) is equal to 1. Since both \(r\) and \((2q+r+1)\) are greater than 1, so they can’t be equal to \(m\) in this case. Again one of \(r, (2q+r+1)\) is odd integer which implies the product \(r(2q+r+1)\) can’t be equal to \(2^l\).  \(\Rightarrow 2^l \neq r(2q+r+1) \) \(\Rightarrow 2^l\) can’t be expressed as the sum of \(r\) consecutive positive integers with \(r>1\) and \(l\ge 1\). Now, \(n=2^0=1 \) also can’t be written in the same manner when \(l=0\). Therefore the positive integers \(n\) that cannot be written as a sum of \(r\) consecutive positive integers, with \(r>1\) , are of the form \(n=2^l\) for some \(l\ge 0\). To show that any number other than of the form \(2^l\) is sum of consecutive integers:
Observe that for \(n=2^l.m\), where \(m\) is an odd number greater than 1, there are two cases:
1. \( m < 2^l \) 
Select \(r\) and \(q\) such that \(r = m\) and \(2q+r+1 = 2^l \). 2. \( m > 2^l \)
Select \(r\) and \(q\) such that \( 2q+r+1 = m\) and \( r = 2^l \). (QED).