Luckyroad21 said:
Your result for n=3 is positive for my conjecture.For n=4, the solutions I obtained from the partial derivatives do not follow the imposed restrictions. For forcing such restrictions on the obtained system I arrived at xi=(2/n)^(1/2) or xi= - (1/2n)^(1/2); it's easy to see that this only adds up to zero if n is a multiple of 3
My crucial conditions are
$$
\dfrac{1}{2}=x_1^2- \sum_{2\leq i < j \leq n} x_ix_j \quad\text{and}\quad x_k\in \left\{x_1\, , \,-\dfrac{x_1}{2}\right\}\quad\text{and}\quad x_1+x_2+\ldots+x_n=0
$$
under the assumption ##x_1\geq x_2 \geq \ldots\geq x_n## and that I didn't make a mistake, especially no sign error anywhere.
Let ##x_k=x_1 \varepsilon_k## with ##\varepsilon_k\in \{1,-1/2\}## and ##\underbrace{x_1 = \ldots = x_1}_{p\text{ times}} >0>\underbrace{-\frac{x_1}{2}=\ldots=-\frac{x_1}{2}}_{q\text{ times}}## and ##p+q=n## these read as
\begin{align*}
\dfrac{1}{2}&=x_1^2\left(1-\sum_{2\leq i < j \leq n} \varepsilon_i\varepsilon_j\right)\quad\text{and}\quad 2p=q=n-p \quad\text{or}\quad n\stackrel{(*)}{=}3p
\end{align*}
with ##1=\varepsilon_1=\varepsilon_2=\ldots=\varepsilon_{p}## and ##-1/2=\varepsilon_{p+1}=\ldots=\varepsilon_n.##
We get for ##n=3## that ##(p,q)\in \{(0,3),(1,2),(2,1),(3,0)\}## where only ##(p,q)=(1,2)## fulfills ##2p=q.## Hence
$$
\dfrac{1}{2}=x_1^2\left(1-\varepsilon_2\varepsilon_3\right)=x_1^2\cdot \dfrac{3}{4}\text{ and }x_1=\sqrt{\dfrac{2}{3}}\, , \,x_2=x_3=-\sqrt{\dfrac{1}{6}}.$$
So all depends on whether (*) has a solution, i.e. whether ##3\,|\,n.##