A finite sum over $\pm 1$ vectors $$\mbox{What is a nice way to show}\quad
\sum_{\vphantom{\Large A}u\ \in\ \left\{-1,+1\right\}^{N}}
\left\vert\,\sum_{i\ =\ 1}^{N}u_{i}\,\right\vert
= N{N \choose N/2}
\quad\mbox{when}\  N\ \mbox{is even ?.}
$$
Could there be a short inductive proof ?.

If there are $r$ -1s in $u$, then $|\sum u_i| = |N-2r|$
Let $M = \frac{N}{2} -1$. Thus we are looking at
$$ \sum_{r=0}^{N} \binom{N}{r} |N - 2r| = 2\sum_{r=0}^{M} \binom{N}{r} (N-2r) $$
(using $\binom{N}{r} = \binom{N}{N-r}$)
Now if $P(x) = \sum_{k=0}^{N} a_k x^k$, then the partial sum $a_o + a_1 + \dots + a_i$ is given by the coefficient of $x^i$ in the series expansion of $\dfrac{P(x)}{1-x}$
Let $$Q(x) = \sum_{r=0}^{N} q_r x^r = \sum_{r=0}^{N} \binom{N}{r}(N-2r)x^r = N\sum_{r=0}^{N} \binom{N}{r}x^r  - \sum_{r=0}^{N} \binom{N}{r}2rx^r =$$
$$= N(1+x)^N - 2xN(1+x)^{N-1} = N(1+x)^{N-1}(1-x)$$
(We used: $(1+x)^N = \sum_{r=0}^{N} \binom{N}{r} x^r$ and then differentiate and multiply by $x$)
We are interested in the partial sums of the coefficients of $Q(x)$: $2(q_0 + q_1 + \dots + q_M)$
Thus we need to look at the coefficient of $x^M$ in $\dfrac{Q(x)}{1-x} = N(1+x)^{N-1}$, which is $N\binom{N-1}{M}$.
Thus your answer is $$ 2N \binom{N-1}{M}$$ which easily simplifies to
$$ N \binom{N}{N/2}$$
(remember, $M = \frac{N}{2} -1$)
