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 ?.

Since $\sum_{i=1}^N|u_i|=||u||_1$ and $u\in \{-1,1\}^N$, we see that there are $\binom{N}{k}$ vectors $u$ which has $k$ $1$'s, $0\le k\le N$. Thus the required sum can be rewritten as $$\sum_{k=0}^N|2k-N|\binom{N}{k}=\sum_{k=0}^{N/2}(N-2k)\binom{N}{k}+\sum_{k= N/2+1}^N(2k-N)\binom{N}{k}\\ =N\binom{N}{N/2}+\sum_{k=N/2+1}^{N}2k \binom{N}{k}-\sum_{k=0}^{N/2} 2k\binom{N}{k}\\=N\binom{N}{N/2}+2N\sum_{k=N/2}^{N-1}\binom{N-1}{k}-2N\sum_{k=0}^{N/2-1}\binom{N-1}{k}$$ It can be easily seen that the last two terms sum to $0$. So the result follows. 
