There is already a fine answer. Here are a few comments and generalizations that might lead to another. Together they are slightly redundant but I don't know which (if any) would work best.
Along with the given sum $$f(n,r)=\sum_{k=0}^{n} (-1)^k \binom{2n}{k} (n-k)^r$$ consider also $$g(n,r)=\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}(n-k)^r.$$
The observation (ignoring the part about odd $r$) is that, seemingly,
$f(n,r)=0$ for even $1 \le r \lt n.$
Here is a claim which looks nicer although it is really no stronger and no weaker:
$g(n,r)=0$ for $0 \le r \lt n.$
The reason the two are the same (aside from the well known case of $r=0$) is that for odd $r,$ $g(n,r)=f(n,r)-f(n,r)$ for odd $r$ and for even $r$ $g(n,r)=f(n,r)+f(n,r)+(-1)^n\binom{2n}{n}0^r.$
However there might be a nice proof for all $r$ which ignores the easier cancellation for odd $r.$
Also, we will note below that "$(n-k)^r$ for $r \lt N$" can be replaced by "$p(k)$ for any polynomial $p(t)$ of degree less than $N.$"
The bounds on $r$ are more restrictive than needed. Computations leave one highly confident that
$f(n,r)=0$ for even $1 \le r \lt 2n.$
Equivalently
$g(n,r)=0$ for $0 \le r \lt 2n.$
of course for the rather trivial reason $g(n,r)=0$ for any odd $r.$
Since the functions $t^0,t^1,\cdots,t^{2n-1}$ are a basis for the space $P_{2n-1}(t)$ of polynomials of degree less than $2n,$ the last claim that
$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}(n-k)^r=0$ for $0 \le r \lt 2n$
is equivalent to
$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}p(n-k)=0$ for all $p \in P_{2n-1}(t)$
and hence also to
$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}q(k)=0$ for all $q \in P_{2n-1}(t)$
This fits in nicely with the remark about finite differences.
Using a different basis
$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}\binom{k}{r}=0$ for $0 \le r \lt 2n.$
Now consider for a polynomial $p=p(t)$ the sum
$$G(N,p)=\sum_{k=0}^{N} (-1)^k \binom{N}{k}p(k).$$
The previous claim
$g(n,r)=G(2n,(n-t)^r)=0$ for $0 \le r \lt 2n$
which , as explained, is equivalent to
$G(2n,p(k))=0$ for any $p\in P_{2n-1}$
Does not depend on the bound being even:
$G(N,p(k))=0$ for any $p \in P_{N-1}$
Finally, here is another perspective which follows from (and imples) some of the above. Consider the polynomials
$$h_{n,r}(t)=\sum_{k=0}^{n} t^k \binom{2n}{k}(n-k)^r.$$
The starting claim was
For even $1 \lt r \lt 2n,$ $h_{n,r}(-1)=0$ i.e. $(t+1)$ divides $h_{n,r}.$
In fact
$(t+1)^{2n-r}$ divides $h_{n,r}.$
Again, the moves of changing to $\sum_{k=0}^{2n} t^k \binom{2n}{k}(n-k)^r$ and then $\sum_{k=0}^{N} t^k \binom{N}{k}p(k)$ are possible with the last thing being divisible by $(t+1)^{N-r}$ where $r \leq N$ is the degree of $p(t).$