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 $p(t).$

(b)). $\endgroup$ – darij grinberg Nov 4 '18 at 2:11