Here is a proof using combinatorial algebra on binomial coefficients.
We seek to verify that with $N\ge 1$ and $0\le n\le N$ we have
$$2^N = \sum_{m=0}^N \sum_{r=0}^n \sum_{s=0}^m
(-1)^{n+m} (-2)^{r+s}
{n\choose r} {m\choose s} {N-r\choose m} {N-s\choose n}.$$
We can re-write the RHS as
$$\sum_{m=0}^N (-1)^{n+m}
\sum_{r=0}^n (-2)^r {n\choose r} {N-r\choose m}
\sum_{s=0}^m (-2)^s
{m\choose s} {N-s\choose n}.$$
Working with the inner sum we get
$$[z^n] (1+z)^N
\sum_{s=0}^m (-2)^s {m\choose s} (1+z)^{-s}
\\ = [z^n] (1+z)^N (1-2/(1+z))^m
= [z^n] (1+z)^{N-m} (z-1)^m.$$
The middle sum yields
$$[w^m] (1+w)^N
\sum_{r=0}^n (-2)^r {n\choose r} (1+w)^{-r}
\\ = [w^m] (1+w)^N (1-2/(1+w))^n
= [w^m] (1+w)^{N-n} (w-1)^n.$$
Combining these in the outer sum we have
$$(-1)^n [z^n] (1+z)^N
\sum_{m=0}^N (-1)^m (1+z)^{-m} (z-1)^m
[w^m] (1+w)^{N-n} (w-1)^n
\\ = (-1)^n [z^n] (1+z)^N
\sum_{m=0}^N (-1)^{N-m} (1+z)^{-N+m} (z-1)^{N-m}
[w^{N-m}] (1+w)^{N-n} (w-1)^n
\\ = (-1)^{n+N} [z^n] (z-1)^N
\sum_{m=0}^N (-1)^m (1+z)^m (z-1)^{-m}
[w^{N-m}] (1+w)^{N-n} (w-1)^n
\\ = (-1)^{n+N} [z^n] (z-1)^N [w^N] (1+w)^{N-n} (w-1)^n
\sum_{m=0}^N (-1)^m (1+z)^m (z-1)^{-m} w^m.$$
We may extend the sum to infinity owing to the coefficient extractor in
$w:$
$$(-1)^{n+N} [z^n] (z-1)^N [w^N] (1+w)^{N-n} (w-1)^n
\frac{1}{1+(1+z)w/(z-1)}
\\ = (-1)^{n+N} [z^n] (z-1)^{N+1} [w^N] (1+w)^{N-n} (w-1)^n
\frac{1}{z-1+(1+z)w}
\\ = (-1)^{n+N+1} [z^n] (z-1)^{N+1} [w^N] (1+w)^{N-n} (w-1)^n
\frac{1}{1-z-w-wz}.$$
The contribution from $w$ is
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{N+1}} (1+w)^{N-n} (w-1)^n \frac{1}{1-z-w-wz}.$$
Now put $w/(1+w) = v$ so that $w=v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$
and $1+w = 1/(1-v)$ to get
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{N+1}} (1-v)
(2v-1)^n \frac{1}{1-z-v/(1-v)-vz/(1-v)} \frac{1}{(1-v)^2}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{N+1}}
(2v-1)^n \frac{1}{(1-z)(1-v)-v-vz}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{N+1}}
(2v-1)^n \frac{1}{1-z-2v}
\\ = (-1)^n \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{N+1}}
(1-2v)^n \frac{1}{1-z-2v}
\\ = (-1)^n \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{N+1}}
(1-2v)^{n-1} \frac{1}{1-z/(1-2v)}.$$
Extracting the coefficient in $z$ we get
$$(-1)^{N+1} \sum_{q=0}^n {N+1\choose q} (-1)^{N+1-q}
\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{N+1}}
(1-2v)^{n-1} \frac{1}{(1-2v)^{n-q}}
\\ = (-1)^{N+1} \sum_{q=0}^n {N+1\choose q} (-1)^{N+1-q}
\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{N+1}}
(1-2v)^{q-1}
\\ = (-1)^{N+1} \sum_{q=0}^n {N+1\choose q} (-1)^{N+1-q}
{q-1\choose N} 2^N (-1)^N
\\ = 2^N (-1)^N \sum_{q=0}^n {N+1\choose q} (-1)^q
{q-1\choose N}
\\ = 2^N + 2^N (-1)^N \sum_{q=1}^n {N+1\choose q} (-1)^q
{q-1\choose N}$$
We almost have the claim. We get for the remaining sum
$$2^N (-1)^N (N+1) \sum_{q=1}^n \frac{1}{q} (-1)^q
{N\choose q-1} {q-1\choose N}.$$
The only non-zero pair of binomial coefficients is when $q=N+1$. But
$q\le n$ and $n\le N$ as per the original problem, so we get zero and
may conclude. We also see that when $n\gt N$ the sum contributes with
one term for a total of
$$2^N + 2^N (-1)^N (N+1) \frac{(-1)^{N+1}}{N+1} = 0.$$
Remark. For the case when $N=0$ there is only one possible value
for $n$, which is zero also. Recall the closed form
$$(-1)^{n+N+1} [z^n] (z-1)^{N+1} [w^N] (1+w)^{N-n} (w-1)^n
\frac{1}{1-z-w-wz}.$$
In this case we are extracting the constant coefficients. We get
starting with $w$
$$(-1)^{n+N+1} [z^n] (z-1)^{N+1} \frac{(-1)^n}{1-z}
= (-1)^{N+1} (-1)^{N+1} = 1.$$
This means that the formula holds for $N=0$ as well.
