Combinatorial 0-1 vector problem Let $M \in \{0, 1\}^{n\times n}$.
Given a constant integer $c \ge 2$, let the number of $1$s in each row be equal to $\frac{n}{c}$ (assuming $c$ is a divisor of $n$). Let $\mathcal{M}_c$ be the set of such matrices.
Given a constant $\beta \in (0,1)$ and a matrix $M \in \mathcal{M}_c$, we say that "a row vector $r$ of $M$ is $\beta$-covered" if the number of row vectors $r'$ of $M$ such that $\,r' r^{\top} \ge \frac{n}{c^2}\,$ is greater than $n^{\beta}$.
Question: When $n$ approaches infinity, what is the minimum number $N$ of $\beta$-covered rows of a matrix in $\mathcal{M}_c$? -- (My conjecture is $N=n-\Theta(n^{\beta})$).
---
PS: My previous questions 0-1 matrix combinatorial problem and A combinatorial 0-1 matrix problem arose from this problem.
 A: Here is an example of $\approx n^{1/3}$ vectors of length $n/c$ with overlaps strictly less than $n/c^2$. Probably, one can do $\approx n$ too, but I do not see any neat construction.
Take a prime $p$ of the form $p=cq-1$. Now, for the index set, use $\mathbb Z_p^2\times\{1,2,\dots,p+1\}$, so $n=p^2(p+1)$. For the $k$-th vector $k=0,1,2,\dots,p-1$, place $1$'s at the triples $(a,b,c)$ with $a+kb\in\{0,1,\dots,q-1\}$ (in $\mathbb Z_p$) and $c\ne k$. Then the cardinality of each set is $p^2q=n/c$. The overlaps are $q^2(p-1)$ (every linear system has unique solution and excluding $2$ indices leaves us with $p-1$ admissible ones). However
$$
\frac {q^2(p-1)}n=\frac{q^2}{p^2}\frac{cq-2}{cq}=\frac{q^2}{p^2}\left(1-\frac 2{cq}\right)<\frac{q^2}{p^2}\left(1-\frac 1{cq}\right)^2=c^{-2}\,.
$$
Small edit
For $c=2$ there is a nice example with $n/2$ vectors when $n=2^s-2$. Take the $s$-dimensional discrete unit cube $Q=\{-1,1\}^s$ and take all Walsh functions $x_I=\prod_{i\in I}x_i$ with odd $|I|$. They are all $1$ at $e_+=(1,\dots,1)$ and $-1$ at $e_-=(-1,\dots,-1)$, so the vectors $v_I=(1+x_I)/2$ considered on $Q\setminus\{e_+,e_-\}$ all have $\frac n2$ ones and intersections of size $\frac{n-2}4$. I guess we can generalize this to other $c$ as well, but it is pretty clear now that if you cannot get your boung from Cauchy-Schwarz in this problem, then it is most likely just not there. 
