The Sum of the Odd Divisors of n 
The sum of the odd divisors of n is $-\sum_{d|n}(-1)^{n/d}d$, and if n is even, then
  $$\sum_{d|n}(-1)^{n/d}d=2\sigma(n/2)-\sigma(n)$$

Could you give me some hints on that?

For the first part, you could try writing $n=2^km$, where $m$ is odd. Then the divisors of $n$ are the numbers $d,2d,4d,\dots,2^kd$ where $d$ runs through the odd divisors of $n$, which is to say, through the divisors of $m$. Now you can rewrite your sum to involve a sum on powers of 2. 
