Let us re-write this as follows:
$${q\choose n} =
(n+1) {n+1+k+q\choose n+1}
\sum_{j=0}^n \frac{(-1)^j}{n+1+k+q-j}
{n\choose j} {2n+k-j\choose n}
\\ = (n+1+k+q) {n+k+q\choose n}
\sum_{j=0}^n \frac{(-1)^j}{n+1+k+q-j}
{n\choose j} {2n+k-j\choose n}.$$
where clearly $q\ge n.$ We have
$${n+k+q\choose n} {n\choose j}
= \frac{(n+k+q)!}{(k+q)! j! (n-j)!}
= {n+k+q\choose j} {n+k+q-j\choose n-j}.$$
Substititute into the identity to get for the RHS
$$\sum_{j=0}^n (-1)^j {n+k+q+1\choose j}
{n+k+q-j\choose n-j} {2n+k-j\choose n}.$$
We put
$${n+k+q-j\choose n-j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+k+q-j}
\; dz$$
and
$${2n+k-j\choose n} =
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} (1+w)^{2n+k-j}
\; dw.$$
Observe that the first of these two vanishes when $j\gt n$ so it
provides range control and we may raise the upper limit of the sum to
$n+k+q+1.$ We get
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} (1+w)^{2n+k}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k+q}
\\ \times \sum_{j=0}^{n+k+q+1} {n+k+q+1\choose j}
(-1)^j \frac{z^j}{(1+z)^j (1+w)^j}
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} (1+w)^{2n+k}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k+q}
\left(1-\frac{z}{(1+z)(1+w)}\right)^{n+k+q+1}
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} \frac{1}{(1+w)^{q-n+1}}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \frac{1}{1+z}
\left(1+w+wz\right)^{n+k+q+1}
\; dz\; dw
.$$
We evaluate this using the fact that residues sum to zero. We get for
the residue at $z=-1$ (flip sign)
$$-(-1)^{n+1}\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} \frac{1}{(1+w)^{q-n+1}}
\; dw
\\ = (-1)^{n}
\times
(-1)^n {q-n+n\choose q-n}
\; dw = {q\choose q-n} = {q\choose n}.$$
This is the target result, so we just need to show that the
contribution from the residue at infinity is zero. We find
$$-\mathrm{Res}_{z=\infty} \frac{1}{z^{n+1}} \frac{1}{1+z}
\left(1+w+wz\right)^{n+k+q+1}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+1} \frac{1}{1+1/z}
\left(1+w+w/z\right)^{n+k+q+1}
\\ = \mathrm{Res}_{z=0} \frac{1}{z} z^{n+1} \frac{1}{1+z}
\frac{1}{z^{n+k+q+1}}
\left(z(1+w)+w\right)^{n+k+q+1}
\\ = \mathrm{Res}_{z=0}
\frac{1}{z^{k+q+1}} \frac{1}{1+z}
\left(z(1+w)+w\right)^{n+k+q+1}.$$
Extracting coefficients yields
$$\sum_{p=0}^{k+q} {n+k+q+1\choose k+q-p}
(1+w)^{k+q-p} w^{n+p+1} (-1)^{p}.$$
On substituting this into the integral in $w$ we obtain
$$[w^n] \frac{1}{(1+w)^{q-n+1}}
\sum_{p=0}^{k+q} {n+k+q+1\choose k+q-p}
(1+w)^{k+q-p} w^{n+p+1} (-1)^{p}
\\ = \sum_{p=0}^{k+q} {n+k+q+1\choose k+q-p}
[w^n] w^{n+p+1} (1+w)^{k+n-p-1} (-1)^{p}.$$
This is zero since
$$[w^n] w^{n+p+1} (1+w)^{k+n-p-1} =
[w^0] w^{p+1} (1+w)^{k+n-p-1} = 0$$
(no constant coefficient in $w$ because $p\ge 0$, we can have
$k+n-p-1$ positive or negative here since there is never a pole at
zero). This concludes the argument. We have shown that the sum indeed
evaluates to ${q\choose n}$ as claimed.
