Combinatorics That Looks Similar to Vandermonde's Identity
Note that $$\sum_{r\ge 0} a_{2r} = \sum_{r\ge 0} \frac{1+(-1)^r}{2}a_r.$$ Taking $$a_r=\binom{k}{r}\binom{n-k}{k-r}$$ yields \begin{align} \sum_{r\ge0} \binom{k}{2r}\binom{n-k}{k-2r} &= \sum_{r\ge0} \frac{1+(-1)^r}{2} \binom{k}{r}\binom{n-k}{k-r} \\ &= \frac{1}{2}\sum_{r\ge0} \binom{k}{r}\binom{n-k}{k-r} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r} \\ &= \frac{1}{2}\binom{n}{k} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r}. \end{align}
I don't know whether the remaining sum can be simplified.