Suppose we have $n$ surfaces lying in unit square on the plane, each of them has area equal to $S.$ What we can say about area of intersection of $k$ of them?
I want find the formula for such maximal number $I(n,S,k),$ for which for every position and form of figures we can find $k$ of them, which intersection area is at least $I(n,S,k).$ Maybe anyone know it?
Since the geometry is irrelevant and you are just considering $n$ subsets of equal measure, $I(n,S,k)$ is also the maximum value such that, given any $n$ measure-$S$ subsets of the unit-circumference circle, there must be $k$ subsets with a mutual intersection of at least $I(n,S,k)$.
One way to minimize maximum mutual overlap is to distribute $n$ arcs/intervals equally around the circle: Going around the circle, start a new arc every $1/n$ units. For small $S$, two consecutive arcs will have an intersection of measure $\max\left(0,S - \frac{1}{n}\right)$. Then $k$ consecutive arcs with have mutual intersection of $\max\left(0,S - \frac{k-1}{n}\right)$ if $S$ isn't too large for wrap-around to be a factor, that is, if $S \leq 1 - \frac{1}{n}$. For $S = 1-\epsilon$ with $0 \leq \epsilon < \frac{1}{n}$, any $k$ of the $n$ arcs have a mutual overlap of $1 - k\epsilon$.
Your function, for $S \in [0,1]$ and integers $2 \leq k \leq n$, is: $$ I(n,S,k) = \left\{ \begin{array}{l l} 0, & S \in \left[0, \frac{k-1}{n}\right]; \\ S - \frac{k-1}{n}, & S \in \left(\frac{k-1}{n} , 1 - \frac{1}{n} \right]; \\ 1 - k(1-S), & S \in \left(1 - \frac{1}{n}, 1 \right]. \end{array}\right. $$
