TITLE: Chips do not form rectangle on board
QUESTION [8 upvotes]: Given is an $n\times n$ board with $n\geq 3$. We place a chip in some cells, so that no four chips form a rectangle with sides parallel to the sides of the board. How many chips can we place, at most?
We can place $2n$ chips in positions $(i,i)$ for $1\leq i\leq n$, $(i,i+1)$ for $1\leq i\leq n-1$, and $(n,1)$. Is it possible to improve?

REPLY [2 votes]: I can always reach $3n-3$. 
Write our disposition of chips as follows : the $n$ columns are written as $S_1,\cdots,S_n$ where $S_i \subseteq \{1,\cdots,n\}$ and the condition that we have no rectangles of chips is equivalent to $|S_i \cap S_j| \le 1$, as in aes's answer ; having a rectangle means there is two columns in which the rectangle appears, and a rectangle appears in these two columns if and only if these two columns both have chips in some pair of rows, which explains the criterion. 
Split in the cases $n$ even and $n$ odd. In the case $n$ even, let 
$$
S_1 = \{ 1,\cdots,n/2 \}, \quad S_2 = \{n/2,\cdots,n\}
$$
and in the case $n$ odd, we pick
$$
S_1 = \{1,\cdots,(n+1)/2\}, \quad S_2 = \{ (n+1)/2,\cdots, n \}. 
$$
We have $S_1 \cap S_2 =\{(n+1)/2\}$ or $\{n/2\}$ depending on whether $n$ odd or even. 
Now we will pick the remaining columns with only two elements, so ensuring that their intersection contains at most one element is equivalent to picking them distinct. But an easy way of ensuring that $S_1 \cap S_k$ and $S_2 \cap S_k$ have at most one element is to build $S_k$ by picking one element in $S_1 \backslash S_2$ and one element in $S_2 \backslash S_1$. This means there are

$n$ even, $|S_1| = n/2$, $|S_2| = n/2+1$, so there are $(n/2-1)(n/2) = \frac{n^2}4 - \frac n2 \ge n-2$ ways of doing this since $(n/2-1)(n/2)-(n-2) = \frac{(n-3)^2 - 1}4 \ge 0$ for $n \ge 4$ (this fits in our assumptions on $n$), 
$n$ odd, $|S_1| = |S_2| = \frac{n+1}2$, so there are $\frac{(n-1)^2}4 \ge n-2$ ways of doing this since $\frac{(n-1)^2}4 -(n-2) = \frac{(n-3)^2}4 \ge 0$. 

Counting the number of chips realized in both cases, we get $3n-3$. 
However, this is sadly far from being generally optimal ; it is not even correct for $n=6$, where I have this arrangement of $16 > 15 = 3 \times 6 - 3$ chips :
$$
\begin{matrix}
 -       & -       & \bullet & \bullet & -       & -       \\
 -       & \bullet & \bullet & -       & \bullet & -       \\
 -       & \bullet & -       & \bullet & -       & \bullet \\
 \bullet & \bullet & -       & -       & -       & -       \\
 \bullet & -       & -       & \bullet & \bullet & -       \\
 \bullet & -       & \bullet & -       & -       & \bullet \\
\end{matrix}
$$ 
I obtained this by trying to fill the first three columns and then "fill in the last columns" as in the case where I restricted my attention to the first two columns and got $3n-3$. This gave me some extra room, but the combinatorics involved in "filling in the last columns" do not look simpler, so the solution does not look very pretty ; however, it suggests that the solution should be a little bit more than linear in $n$. The asymptotic solution $n \sqrt n$ of aes doesn't sound bad. 
Hope that helps,