partial alternating sum involving binomial coefficients I came across the following alternating sum
$$
\sum_{k=0}^n (-1)^k \binom{2n}{k} (n-k)^r,\quad 1\leq r < n.
$$
It seems that when $r$ is an even integer the sum is $0$ and when $r$ is an odd integer the sum is not zero (regardless of the parity of $n$). 
[Edited] The case when $r$ is even is easy by symmetry as Darij Grinberg pointed out below. So the question left is how to show that the sum is nonzero when $r$ is odd.
The main difficulty I have proving this is that the sum is only from $0$ to $n$ instead of to $2n$, and I don't see how to apply the classical methods such as finite differences, integral representation, series, etc.
 A: In a post on Math Stack Exchange, MSE 2827591, I prove the following:
$$\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s =
\binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}.
$$ 
The OP's formula can be put in the form on the LHS of this equation.  By inspection the questions of concern can be answered; namely, the value of zero for $s$ an even integer, and non-zero otherwise.
A: Here's a very clunky approach that might not be close to a general solution. However, it is a proof for $r\le 99$. Define $p_r(n)$ by
$$ \sum_{k=0}^n (-1)^k \binom {2n}k (n-k)^r =
   (-1)^n\binom{2n}{n}\frac{n^2\,p_r(n)}{2\prod_{t=0}^{(r-1)/2} (2n-2t+1)}. $$
For $r=3,5,\ldots,99$, $p_r(n)$ is a polynomial of degree $\frac{r-3}2$ with integer coefficients.
For example $p_3(n) = 1$, $p_5(n)=-4n+1$, $p_7(n)=34n^2-24n+5$.
For odd $r\le 99$, $p_r(n)$ is irreducible, but that is likely to be very hard to prove (or false) for all $r$. A simpler observation is that, for $r\le 99$, all the coefficients of $p_r(n)$ are even, except that the constant term is odd.  Therefore, $p_r(n)$ is an odd integer when $n$ is an integer.
Without caring about its coefficients, maybe there is a direct way to show that $p_r(n)$ is odd whenever $r$ is odd.
A: Just for the curious mind.
Let's suppose $r\rightarrow 2r+1$. Then, the constant term in Brendan McKay's experiment can be given as follows: 
$$\frac{2(-1)^{r-1}(2r+1)!B(2r)}{r!\,2^r}$$
where $B(n)$ are the Bernoulli numbers.
A: 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 $p(t).$
