Given $A\in\{0,1\}^{n\times n}$ with $\operatorname{rank}(A)=r=2^{O((\log\_2n)^{\frac{1}{c+1}})}$.


Denote $\mathsf{1\_n}\in\{0,1\}^n$ as vector with $1$s.


Does
$$\lim\_{n\rightarrow\infty}\mathsf{P\_{A\in\{0,1\}^{n\times n}}}\Bigg(\mathsf{1\_n}'A\mathsf{1\_n}>\frac{(r-1)r^{(\log\_2r)^c}}{2\log\_2r}\Bigg)=1\quad?$$


Infact is there a matrix (family) such that $$\mathsf{1\_n}'A\mathsf{1\_n}<\frac{(r-1)r^{(\log\_2r)^c}}{2\log\_2r}\quad?$$


$c\geq1$ fixed.



Probabilistic statement on matrix ranks
Consider the set of $n\times n$-matrices with entries in $\{0,1\}$ which have at most $r$ distinct rows. The number of such matrices is $2^{rn}r^n$. As long as $n$ and $n-r$ tend to infinity, we have that such a matrix almost surely has rank $r$, and contains close to $n^2/2$ entries equal to 1. 


Next the number of $n\times n$-matrices with all entries 0 with $N$ exceptions is $\binom{n^2}{N}\leq n^{2N}$. Hence the probability that a random matrix with rank $r$ has at most $N$ entries 1 is bounded above by $(1+o(1))\frac{n^{2N}}{2^{rn}r^n}$, which tends to 0 provided that $N\leq\frac{rn\log 2}{2\log n}$. We obtain that almost all random matrices with rank $r$ have more than $\frac{rn\log 2}{2\log n}$ entries equal to 1.


For the second question just take diagonal matrices.


