The complete answer is that, $x_i$ maximizes the given function, if and only if $\sum_i x_i=0$. Here is a proof base on achille hui's great comment:
$$
\sum\limits_{i<j}\|x_i - x_j\|^2 = \frac12\sum\limits_{i,j}\|x_i - x_j\|^2 = \frac12\big({\small 4\sum\limits_j \|x_j\|^2 + 4\sum\limits_i \|x_i\|^2 - 2\langle \sum\limits_i x_i , \sum\limits_j x_j}\rangle\big) $$
$$=16-\|\sum_i x_i\|^2.
$$
In particular, any regular tetrahedron whose vertices lie on the sphere satisfies this condition, but there are other examples, such as a square on the equator.
Edit 1:
It is worth noting that the proof above does not use in any essential way the dimension of the sphere (here $2$), or the number of points (here $4$).
Indeed, if we want to place $N$ points $x_1,\dots,x_N$ on the $n$-dimensional sphere $\mathbb{S}^n \subseteq \mathbb{R}^{n+1}$, maximizing $\sum\limits_{i<j}\|x_i - x_j\|^2$, then the same calculation gives:
$$
\sum\limits_{i<j}\|x_i - x_j\|^2 = \frac12\sum\limits_{i,j}\|x_i - x_j\|^2 =\frac12\sum\limits_{i,j}\|x_i\|^2+\|x_j\|^2-2 \langle x_i , x_j\rangle
$$
Since $$\sum\limits_{i,j}\|x_i\|^2=\sum_i \sum_j \|x_i\|^2=\sum_i N \|x_i\|^2=N \sum_i \|x_i\|^2=N^2
$$
and
$$
\sum\limits_{i,j} \langle x_i , x_j\rangle=\langle \sum\limits_i x_i , \sum\limits_j x_j\rangle,
$$
we get
$$
\sum\limits_{i<j}\|x_i - x_j\|^2=\frac12 \big( 2N^2 - 2\langle \sum\limits_i x_i , \sum\limits_j x_j\rangle\big)
=N^2-\|\sum_i x_i\|^2.
$$
Edit 2: (Generalization)
Here we again focus only on four points:
$f : (0,4] \to (0,\infty)$ be any monotone increasing concave function.
We use $\sum_{i<j} |x_i - x_j|^2
=16 - \left|\sum_i x_i\right|^2.$
Since $f$ is concave and monotone increasing,
$$\sum_{i<j}f\left(|x_i - x_j|^2\right) \le 6f\left(\frac16\sum_{i<j}|x_i-x_j|^2\right) = 6f\left(\frac83-\frac16|\sum_ix_i|^2\right)\le 6f\left(\frac83\right).$$
In particular, $E_f=\sum_{i<j}f\left(|x_i - x_j|^2\right)$ is maximized whenever $|x_i-x_j|$ are all equal, i.e. $x_i$ is a tetrahedron. (This automatically implies $\sum_i x_i=0$).
When $f$ is strictly concave, the tetrahedron is the unique maximizer.
In particular this works for the map $x \to \sqrt{x}$.
Best Answer
Consider these four points:
$$ (0.8,0.6,0) \\ (0.8,-0.6,0) \\ (-0.8,0,0.6) \\ (-0.8,0,-0.6) $$