A simple finite combinatorial sum I found, that seems to work, would have good reasons to work, but I can't find in the literature. I was doing a consistency check for some calculations I'm performing for my master thesis (roughly - about a problem in discrete bayesian model selection) - and it turns out that my choice of priors is only consistent if this identity is true:
$$\sum_{k=0}^{N}\left[ \binom{k+j}{j}\binom{N-k+j}{j}\right] = \binom{N+2j+1}{2j+1}$$
Now, this seems to work for small numbers, but I searched for it a lot in the literature and I can't find it.(I have a physics background so probably my knowledge of proper "literature" is the problem here). Neither I can demonstrate it!
Has anyone of you seen this before? Can this be rewritten equivalently in some more commonly seen way? Can it be proven right...or wrong?
Thanks in advance! :)

We have 
$$\sum_{k=0}^N {k+j\choose j} {N-k+j\choose j}
= \sum_{k=0}^N {k+j\choose j} {N-k+j\choose N-k}
\\ = \sum_{k=0}^N {k+j\choose j} [z^{N-k}] (1+z)^{N-k+j}
= [z^N] (1+z)^{N+j} \sum_{k=0}^N {k+j\choose j} z^k (1+z)^{-k}.$$
Now  we may  extend $k$  beyond  $N$ as  there is  no contribution  to
$[z^N]$  at the  front in  this case  (we have  $z^k (1+z)^{-k}  = z^k
+\cdots$)
$$[z^N] (1+z)^{N+j} \sum_{k\ge 0} {k+j\choose j} z^k (1+z)^{-k}
= [z^N] (1+z)^{N+j} \frac{1}{(1-z/(1+z))^{j+1}}
\\ = [z^N] (1+z)^{N+j} \frac{(1+z)^{j+1}}{(1+z-z)^{j+1}}
= [z^N] (1+z)^{N+2j+1} = {N+2j+1\choose 2j+1}.$$
