TITLE: Combinatorial proof that $\sum \limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$ when $n$ is even
QUESTION [73 upvotes]: In my answer here I prove, using generating functions, a statement equivalent to 
$$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$
when $n$ is even.  (Clearly the sum is $0$ when $n$ is odd.)  The nice expression on the right-hand side indicates that there should be a pretty combinatorial proof of this statement.  The proof should start by associating objects with even parity and objects with odd parity counted by the left-hand side.  The number of leftover (unassociated) objects should have even parity and should "obviously" be $2^n \binom{n}{n/2}$.  I'm having trouble finding such a proof, though.  So, my question is

Can someone produce a combinatorial proof that, for even $n$, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}?$$

Some thoughts so far: 
Combinatorial proofs for $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$ are given by Phira here and by Brian M. Scott here.  The proofs are basically equivalent.  In Phira's argument, both sides count the number of paths of length $2n$ starting from $(0,0)$ using steps of $(1,1)$ and $(1,-1)$.  By conditioning on the largest value of $2k$ for which a particular path returns to the horizontal axis at $(2k,0)$ and using the facts that there are $\binom{2k}{k}$ paths from $(0,0)$ to $(2k,0)$ and $\binom{2n-2k}{n-k}$ paths of length $2n-2k$ that start at the horizontal axis but never return to the axis we obtain the left-hand side.
With these interpretations of the central binomial coefficients $2^n \binom{n}{n/2}$ could count (1) paths that do not return to the horizontal axis by the path's halfway point of $(n,0)$, or (2) paths that touch the point $(n,0)$.  But I haven't been able to construct the association that makes these the leftover paths (nor do all of these paths have even parity anyway).  So perhaps there's some other interpretation of $2^n \binom{n}{n/2}$ as the number of leftover paths.

Update. Some more thoughts:
There's another way to view the identity $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$.  Both sides count the number of lattice paths of length $n$ when north, south, east, and west steps are allowed.  The right side is obvious. The left side has a similar interpretation as before: $\binom{2k}{k}$ counts the number of NSEW lattice paths of length $k$ that end on the line $y=0$, and $\binom{2n-2k}{n-k}$ counts the number of NSEW lattice paths of length $n-k$ that never return to the line $y =0$.  So far, this isn't much different as before.  However, $2^n \binom{n}{n/2}$ has an intriguing interpretation: It counts the number of NSEW lattice paths that end on the diagonal $y = x$ (or, equivalently, $y = -x$).  So maybe there's an involution that leaves these as the leftover paths.  (Proofs of all of these claims can be found on this blog post, for those who are interested.)

REPLY [8 votes]: My alternative solution can be found here (in Section 4), by counting paths: http://arxiv.org/abs/1204.5923