This is not quite a solution of the original problem, but rather of a related and, perhaps, even more natural one:
How small can the ranks of real matrices $B=(b_{ij})_{1\le i,j\le n}$ and $C=(c_{ij})_{1\le i,j\le n}$ be, given that $b_{ii}c_{ii}\ne 0$, while $b_{ij}c_{ji}=0$ whenever $i\ne j$?
The inequality $\rk(B\circ C)\le\rk(B)\rk(C)$ gives
$$\rk(B)\rk(C)\ge n,$$
and I claim that this estimate is sharp. To see this, associate the rows and columns of our matrices with the elements of a finite abelian group $G$ of order $|G|=n$, consider a decomposition $G=G_1\oplus G_2$ and, denoting by $I_1$ and $I_2$ the indicator functions $G_1$ and $G_2$, respectively, for $u,v\in G$ let $b_{uv}:=I_1(u-v)$ and $c_{uv}:=I_2(u-v)$. We have then $b_{uv}c_{vu}\ne 0$ if and only if $u-v\in G_1$ and $v-u\in G_2$, which is equivalent to $u=v$. On the other hand, each row of $B$ is then identical to $|G_1|$ other rows, showing that $\rk(B)\le|G|/|G_1|=|G_2|$ and, similarly, $\rk(C)\le|G_1|$.
I still wonder whether it is possible to modify somehow the constructions to have $B=C$ and $\rk(B)=\rk(C)\approx\sqrt n$.