This sum can be expressed in terms of $2n$-th [forward difference](https://en.wikipedia.org/wiki/Forward_difference):
\begin{split}
\sum_{k=1}^n (-1)^k\frac{k^{p}}{n+k}\binom{2n-1}{n-k} &=
\frac1{2n} \sum_{k=1}^n (-1)^k k^{p} \binom{2n}{n+k} \\
&\mathop{=}_{p\ \text{even}}\frac{(-1)^n}{4n} \sum_{k=0}^{2n} (-1)^k (k-n)^{p} \binom{2n}{k} \\
&=\frac{(-1)^n}{4n}\left.\Delta^{2n}x^p\right|_{x=-n},
\end{split}
which is zero for $p<2n$.