Let's define for every pair of vectors $u,v\in\mathbb{R}^n$, a quantity as follows:
$$f(u,v) = \sum\_{1\leq i,j\leq n}|u\_iu\_j-v\_iv\_j|.$$


I want to find:
$$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$


An easy estimate using the triangle inequality gives $M(n) <2n$, but it seems there should be better upper bounds. 


**Remark**. 1) Let $a = \cos(\pi/8), b = \sin(\pi/8)$, then if for even numbers $n$ we define the vectors $u = \sqrt{\frac{2}{n}}(a,b,a,b,\dots)$ and $ v = \sqrt{\frac{2}{n}}(b,-a,b,-a,\dots)$, we have $f(u,v)=\sqrt{2}n\leq M(n)$.
2)A similar construction shows that for every positive integers $n,k$, $\frac{M(n)}{n}\leq \frac{M(nk)}{nk}.$ 



Maximum of a quantity for two normal orthogonal vectors in $\mathbb{R}^n$
Just use Cauchy-Schwarz and the identity
$$
\sum\_{i,j}(u\_iu\_j-v\_iv\_j)^2=\left[\sum\_i u\_i^2\right]^2+\left[\sum\_i v\_i^2\right]^2-2\left[\sum\_i u\_iv\_i\right]^2=2
$$
to get $f(u,v)\le \sqrt2 n$ for all $n$. As you have observed yourself, this is sharp for even $n$. For odd $n$ you can, probably, do a bit better but the estimates get messy and it is not clear (to me) what the sharp bound really is.


