A subset of unit square 
A subset $\mathcal{S}$ of an unit square is equal to the sum of some
  number of disjoint and congruent squares. The total area of these
  squares is equal to $\frac{17}{50}$. Prove that one can find two
  points in the subset $\mathcal{S}$ such that the distance between them
  is equal to $\frac{1}{2003}$?

Frankly, where to start? I expect this to be sort of problem one can to using pigeonhole principle, but where to find it here? My idea was to consider that the subset is made of $n$ squares of size $a\times a$. Then we have that $na^2=\frac{17}{50}$, but what next?
 A: Work in progress. This seems overly complicated but could provide a suitable route.

The actual target distance given here, $\frac {1}{2003}$, is not important, provided it is smaller than the distance that would produce a border region of less than $1/50$th the unit square area (so any value less than say $\frac{1}{200}$ would be suitable for this proof). Consider such a distance $t$. 
Firstly consider how small a single square must be to exclude having two points $t$ apart within its boundary. The distance across the diagonal of a square is $s\sqrt 2$, with $s$ as its side length. Then $s\sqrt 2 < t$ gives $s<\frac{t}{\sqrt 2}$. Thus we can consider squares below this dimension.
Now consider image of such a square displaced in all possible directions, creating a ring around the square . This is the area that other squares cannot overlap in order to avoid points being distance $t$ apart. For example there are non-overlapping images of the square directly out from all four sides and all four corners. 

For the purposes of this proof consider the $6$ images of the square at corners of a regular hexagon:

This part of the overlap zone can be shared by $12$ other square, $6$ sharing only one square image and $6$ sharing two.
