Using Noam Elkies' answer, we can wlog assume that the entries of $x$ are:

$$ k \textrm{ copies of } \dfrac{k - n}{\sqrt{kn(n-k)}} $$

$$ n - k \textrm{ copies of } \dfrac{k}{\sqrt{kn(n-k)}} $$

and similarly that the entries of $y$ are:

$$ l \textrm{ copies of } \dfrac{l - n}{\sqrt{ln(n-l)}} $$

$$ n - l \textrm{ copies of } \dfrac{l}{\sqrt{ln(n-l)}} $$

where wlog $1 \leq k, l \leq \frac{n}{2}$ (otherwise take complements).

We shall denote the sets of negative coordinates of $x$ and $y$, respectively, by $K$ and $L$ (so $|K| = k$ and $|L| = l$). Let their complements be $K'$ and $L'$.

Now, a pair of indices $i, j$ contributes a nonzero term to the objective if and only if exactly one of $i, j$ is in $K$, and exactly one of $i, j$ is in $L$.

The total number of such pairs (which we can view as edges in the intersection of two complete bipartite graphs) is given by:

$$ | K \cap L | | K' \cap L' | + | K \cap L' | | K' \cap L | $$

If we let $m := | K \cap L |$, this is just:

$$ m (n + m - k - l) + (k - m)(l - m) $$

This is a quadratic function of $m$ with minimiser $\frac{1}{4}(2k + 2l - n)$.

**Case 1:** If $2k + 2l \leq n$, this is minimised on the boundary when $m = 0$. We can take $K$ and $L$ to be disjoint and everything is much simpler. The number of nonzero terms is $kl$, and the size of each term is:

$$ \dfrac{n^2}{\sqrt{lkn^2(n-l)(n-k)}} $$

so the total value of the objective function is:

$$ n \sqrt{\dfrac{kl}{(n - l)(n - k)}} $$

It is now straightforward to see that we want $l$ and $k$ to be as small as possible, namely $1$, giving Strickland's solution as the unique optimum.

**Case 2:** If $n < 2k + 2l$, then the optimum $m = \lfloor \frac{1}{4}(2k + 2l - n) \rfloor$ is still small, in that we have $m \leq \frac{k}{2}$ and $m \leq \frac{l}{2}$. This implies that the number of nonzero pairs is still at least $\frac{1}{4} k l$, so if $kl \geq 4$ then we do worse than Strickland's solution in Case 1.

In the remaining subcase where $kl < 4$ and $n < 2k + 2l$, we necessarily have $n \leq 7$. But all of these small cases have been checked by the OP, so the result holds in general.