<s>[INST] show this sum $\sum_{i=0}^{n}(-1)^{i-j}\binom{n+i}{i}\binom{n+k-i}{k-i}\binom{2n}{n+j-i}=\binom{2n}{n}$ 
Let $n,k$ be positive integers.  Show that
$$\sum_{i=0}^{n}(-1)^{i-j}\binom{n+i}{i}\binom{n+k-i}{k-i}\binom{2n}{n+j-i}=\binom{2n}{n}$$
for all $j=0,1,2,\cdots,k$.

I don't know if this is an existing conclusion. After thinking for a long time, I found it difficult to deal with it.

Batominovski's edit:
However, when $k=0$ and $j=0$, the required sum is
$$S=\sum_{i=0}^n(-1)^i\binom{n+i}{i}\binom{n-i}{-i}\binom{2n}{n-i}=\binom{n}{0}\binom{n}{0}\binom{2n}{n}=\binom{2n}{n}.$$
When $k=1$ and $j=0$, the required sum is
$$S=\binom{n}{0}\binom{n+1}{1}\binom{2n}{n}-\binom{n+1}{1}\binom{n}{0}\binom{2n}{n-1}=(n+1)\binom{2n}{n}-(n+1)\binom{2n}{n-1}.$$
Because $\binom{2n}{n-1}=\binom{2n}{n+1}$, we have
$$(n+1)\binom{2n}{n-1}=\binom{n+1}{1}\binom{2n}{n+1}=\binom{n}{1}\binom{2n}{n}=n\binom{2n}{n}.$$
Hence, $S=(n+1)\binom{2n}{n}-n\binom{2n}{n}=\binom{2n}{n}$.  When $k=1$ and $j=1$, the required sum is
$$S=-\binom{n}{0}\binom{n+1}{1}\binom{2n}{n+1}+\binom{n+1}{1}\binom{n}{0}\binom{2n}{n}=-(n+1)\binom{2n}{n+1}+(n+1)\binom{2n}{n}.$$
So, similar to the previous case, $S=\binom{2n}{n}$.  How to prove this for generalized pairs $(k,j)$?  Is there a combinatorial proof?
 [/INST] Rewrite the binomial coefficients in terms of factorials
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \frac{(n+i)!}{\color{red}{n!}i!} \frac{(n+k-i)!}{\color{red}{n!}(k-i)!} \frac{\color{red}{(2n)!}}{(n+j-i)!(n+i-j)!} =\color{red}{\frac{ (2n)!}{(n!)^2}}.
\end{eqnarray*}
$\color{red}{\text{Cancel}}$ and $\color{blue}{\text{create}}$
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \frac{(n+i)!}{(n+i-j)!\color{blue}{j!}} \frac{(n+k-i)!}{(n+j-i)!\color{blue}{(k-j)!}} \frac{\color{blue}{k!}}{i!(k-i)!} = \color{blue}{\frac{k!}{j!(k-j)!}}.
\end{eqnarray*}
Thus, it suffices to show
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i} \binom{n+i}{j} \binom{n+k-i}{k-j} =\binom{k}{j}.
\end{eqnarray*}
We shall use $2$ coefficient extractors
\begin{eqnarray*}
\binom{n+i}{j} &=& [x^j]: (1+x)^{n+i} \\
\binom{n+k-i}{k-j} &=& [y^{k-j}]:(1+y)^{n+k-i}.
\end{eqnarray*}
So
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i} \binom{n+i}{j} \binom{n+k-i}{k-j} 
&=& \sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i} [x^j]: (1+x)^{n+i} [y^{k-j}]:(1+y)^{n+k-i}\\
&=&[x^j][y^{k-j}]: \sum_{i=0}^{n} (-1)^{i-j} \binom{k}{i}  (1+x)^{n+i} (1+y)^{n+k-i}\\
&=&[x^j][y^{k-j}]: (-1)^j \left( 1-\frac{1+x}{1+y} \right)^k  (1+x)^{n} (1+y)^{n+k}\\
&=&[x^j][y^{k-j}]:(-1)^j  \left( y-x \right)^k  (1+x)^{n} (1+y)^{n}\\
&=& \binom{k}{j}. \\
\end{eqnarray*}
</s>