Notice that this sum can be expressed in terms of $2n$-th 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}}\frac1{4n} \sum_{k=0}^{2n} (-1)^k (k-n)^{p} \binom{2n}{k} \\ &=\frac1{4n}\left.\Delta^{2n}x^p\right|_{x=-n}, \end{split} which is zero for $p<2n$.

This sum can be expressed in terms of $2n$-th 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 $p<2n$.

Notice that this sum equals $$\frac1{2n} \sum_{k=1}^n (-1)^k k^{p} \binom{2n}{n+k}=\frac1{4n} \sum_{k=0}^{2n} (-1)^k (k-n)^{p} \binom{2n}{k}=\frac1{4n}\left.\Delta^{2n}x^p\right|_{x=-n},$$ which is zero for even $p<2n$.

Notice that this sum can be expressed in terms of $2n$-th 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}}\frac1{4n} \sum_{k=0}^{2n} (-1)^k (k-n)^{p} \binom{2n}{k} \\ &=\frac1{4n}\left.\Delta^{2n}x^p\right|_{x=-n}, \end{split} which is zero for $p<2n$.