Determine all positive integer values of (n) such that

$$ { n \choose 0 } + { n \choose 1 } + { n \choose 2 } + { n \choose 3 } \ \bigg| \ 2 ^ { 2008 } $$

What is the sum of all these values? 

CURRENT PROGRESS:

I was able to find out that this is equivalent to $(n+1)(n^2 - n + 6) \,| \,3\times (2)^{2009}$ and opened in 2 cases, $n+1 = 2^a$ and $n+1 = 3\times 2^a$, trying to solve like: $n = 2^a - 1$, then $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8$, doing the same to the 2nd case but couldn't find solutions. Something that should be useful is that $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8 | \space\space 3\times2^{2019-a}$, it has a factor 3 in it.