jmc

The expression $2n-1$ is odd for every integer $n$, and conversely every odd integer takes the form $2n-1$ for some integer $n$. Therefore, there is one solution $n$ for each (not necessarily positive) odd divisor of 20. The positive odd divisors of 20 are 1 and 5, so we solve $2n-1=-5$, $2n-1=-1$, $2n-1=1$, and $2n-1=5$ to find the solutions $n=-2$, $n=0$, $n=1$, and $n=3$. These values for $n$ sum to $\boxed{2}$.