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?

Numerical evidence indicates that the upper limit should be $k$ and not 
$n$ for this to hold including when $n\lt k.$ So we seek to show that
$$\sum_{q=0}^k (-1)^{q-j} {n+q\choose q}
{n+k-q\choose k-q} {2n\choose n+j-q} = {2n\choose n}$$
where $0\le j\le k.$ The LHS is
$$(-1)^j [w^{n+j}] (1+w)^{2n}
\sum_{q=0}^k (-1)^{q} {n+q\choose q} w^q
{n+k-q\choose k-q}
\\ = (-1)^j [w^{n+j}] (1+w)^{2n} 
[z^k] \frac{1}{(1+wz)^{n+1}} \frac{1}{(1-z)^{n+1}}.$$
The inner term is
$$\mathrm{Res}_{z=0} \frac{1}{z^{k+1}} 
\frac{1}{(1+wz)^{n+1}} \frac{1}{(1-z)^{n+1}}.$$
Residues sum to zero and the residue at infinity is zero by inspection. 
We get for the residue at $z=1$
$$(-1)^{n+1} \mathrm{Res}_{z=1} \frac{1}{z^{k+1}} 
\frac{1}{(1+wz)^{n+1}} \frac{1}{(z-1)^{n+1}}
\\ = (-1)^{n+1} \mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{k+1}} 
\frac{1}{(1+w+w(z-1))^{n+1}} \frac{1}{(z-1)^{n+1}}
\\ = \frac{(-1)^{n+1}}{(1+w)^{n+1}} 
\mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{k+1}} 
\frac{1}{(1+w(z-1)/(1+w))^{n+1}} \frac{1}{(z-1)^{n+1}}
\\ = \frac{(-1)^{n+1}}{(1+w)^{n+1}} 
\sum_{q=0}^n {n+q\choose q} (-1)^q \frac{w^q}{(1+w)^q}
(-1)^{n-q} {k+n-q\choose k}
\\ = - \sum_{q=0}^n 
{n+q\choose q} \frac{w^q}{(1+w)^{n+1+q}} {k+n-q\choose k}.$$
Substitute into the coefficient extractor in $w$ to get
$$- (-1)^j \sum_{q=0}^n 
{n+q\choose q} {k+n-q\choose k}
[w^{n+j-q}] (1+w)^{n-1-q}.$$
Now with $0\le q\le n-1$ and $j\ge 0$ we have $[w^{n+j-q}] 
(1+w)^{n-1-q} = 0.$ This leaves $q=n$ which yields
$$-(-1)^j {2n\choose n} {k\choose k} [w^j] \frac{1}{1+w}
= - {2n\choose n}.$$
This is the claim. We have the result if we can show
that the residue at $z=-1/w$ makes for a zero contribution.
We get
$$\frac{1}{w^{n+1}} \mathrm{Res}_{z=-1/w} \frac{1}{z^{k+1}} 
\frac{1}{(z+1/w)^{n+1}} \frac{1}{(1-z)^{n+1}}.$$
This requires
$$\frac{1}{n!} 
\left(\frac{1}{z^{k+1}} \frac{1}{(1-z)^{n+1}}\right)^{(n)}
= \frac{1}{n!} \sum_{q=0}^n {n\choose q}
\frac{(-1)^q (k+q)!}{z^{k+1+q} \times k!}
\frac{(n+n-q)!}{(1-z)^{n+1+n-q} \times n!}
\\ = \sum_{q=0}^n {k+q\choose k} (-1)^q \frac{1}{z^{k+1+q}}
{2n-q\choose n} \frac{1}{(1-z)^{2n+1-q}}.$$
Evaluate at $z=-1/w$ and restore the factor in front:
$$\frac{1}{w^{n+1}}
\sum_{q=0}^n {k+q\choose k} (-1)^{k+1} w^{k+1+q}
{2n-q\choose n} \frac{1}{(1+1/w)^{2n+1-q}}.$$
Applying the coefficient extractor in $w$ we get
$$(-1)^j [w^{n+j}] (1+w)^{2n}
\frac{1}{w^{n+1}} w^{k+1+q} \frac{w^{2n+1-q}}{(1+w)^{2n+1-q}}
\\ = (-1)^j [w^{n+j}] (1+w)^{q-1} w^{n+k+1}
= (-1)^j [w^j] (1+w)^{q-1} w^{k+1} = 0$$
because $j\le k.$ This concludes the argument.
