Consider lattice paths consisting of $2n$ steps, each of which is either $(1,1)$ or $(1,-1)$. The number of such lattice paths that return to the horizontal axis only at times that are a multiple of $4$ is given by $2^n \binom{n}{n/2}$. Can someone provide a combinatorial proof of this fact?

Background: A few months ago on math.SE I asked for a combinatorial proof of the identity $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2},$$ when $n$ is even.

The non-alternating version is $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} = 4^n.$$ There are several combinatorial proofs of the non-alternating version, and I hoped to adapt one of them. One such proof is that $\binom{2k}{k} \binom{2n-2k}{n-k}$ counts the number of paths of length $2n$ with $2k$ steps above the horizontal axis and $2n-2k$ steps below it. Summing up over all values of $k$ gives the total number of paths of length $2n$, which is $2^{2n} = 4^n$. (I believe I saw this argument in Feller's *An Introduction to Probability Theory and Its Applications*. It's also in this note by David Callan.)

If we take the alternating version, the paths with positive parity are those with $2k$ steps above the axis for $k$ even, and the paths with negative parity are those with $2k$ steps above the axis for $k$ odd. For each path, break it every time it returns to the horizontal axis. This partitions each path into a number of segments equal to the number of times it returns to the axis. For every path that has a last segment consisting of $2j$ steps for $j$ odd, flip this segment over the horizontal axis. This mapping is an involution and changes the path's parity. Since every odd-parity path must have at least one such odd segment, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k$$ must count the number of paths that have no odd segment; i.e., the number of paths whose returns to the horizontal axis occur only at multiples of $4$. If $n$ is odd, there are no such paths without an odd segment, but if $n$ is even, there are apparently $2^n \binom{n}{n/2}$ of them.

However, I was unable to find an independent combinatorial proof that $2^n \binom{n}{n/2}$ counts the number of lattice paths of length $2n$ with no odd segments. (Again, an "odd" segment here is one of length $2j$, where $j$ is odd.) The math.SE question remained unanswered for over two months until I found a different way to prove the identity I was after combinatorially, but this other way doesn't involve lattice paths. After all the time I spent trying the lattice path approach I would like to see an independent combinatorial proof that $2^n \binom{n}{n/2}$ counts the number of lattice paths whose return times to the axis are multiples of $4$.