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.

Your identity is correct. I don't know a reference offhand, but here is a proof.

The right side, $\binom{N+2j+1}{2j+1}$, is the number of bitstrings of length $N+2j+1$ consisting of $N$ zeroes and $2j+1$ ones.

The sum on the left counts the same set of bitstrings. Namely, for $0\le k\le N$, the term $\binom{k+j}j\binom{N-k+j}j$ is the number of those bitstrings in which the middle one, with $j$ ones on either side, is in the $k+j+1^\text{st}$ position; i.e., it has $k$ zeroes and $j$ ones to the left, $N-k$ zeroes and $j$ ones to the right.

P.S. I found your identity in László Lovász, Combinatorial Problems and Exercises, North-Holland, 1979 (the first edition), where it is Exercise 1.42(i) on p. 18, with hint on p. 96 and solution on p. 172. Lovász gives the identity in the following (more general) form: $$\sum_{k=0}^m\binom{u+k}k\binom{v-k}{m-k}=\binom{u+v+1}m.$$ If we set $m=N$, $u=j$, $v=N+j$, this becomes $$\sum_{k=0}^N\binom{j+k}k\binom{N+j-k}{N-k}=\binom{N+2j+1}N$$ which is plainly equivalent to your identity $$\sum_{k=0}^N\binom{k+j}j\binom{N-k+j}j=\binom{N+2j+1}{2j+1}.$$

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 $z^k (1+z)^{-k} = z^k +\cdots$)

$$[z^N] (1+z)^{N+j} \sum_{k\ge 0} {k+j\choose j} z^k (1+z)^{-k} = [z^N] (1+z)^{N+j} \frac{1}{(1-z/(1+z))^{j+1}} \\ = [z^N] (1+z)^{N+j} \frac{(1+z)^{j+1}}{(1+z-z)^{j+1}} = [z^N] (1+z)^{N+2j+1} = {N+2j+1\choose 2j+1}.$$