Marking Squares on a chessboard

Assume having grid with dimension 2014 x 2014, so there are separated 106 x 106 sub-grids. So it has at least 21 x 106 x 106 marked unit squares. The problem becomes to find solution to distribute marked unit squares as much as possible in border of grid 2014 x 2014. Then when cut one border row and one column to get 2013 x 2013 grid with minimal possible number of unit squares. By some particular trials to get the maximum possible number of marked unit squares that can be cut: $22 \times 106 - 1 = 2331$ The minimal possible number of unit squares is: $21\times 106^2 - 2331 = \boxed{233625}$

The problem becomes much easier when you use modular arithmetic. Consider a 19x19 subgrid: it is composed of the points $(x,y) \in \{a,a+1,a+2...,a+18\} \times \{b,b+1,b+2...,b+18\}$ . Now consider the map $V: (a,b) \rightarrow (n,m)$ where $(a,b) \equiv (n,m) \pmod{19}$ and $1 \leq n,m \leq 19$ . $V$ maps any subgird $(x,y) \in \{a,a+1,a+2...,a+18\} \times \{b,b+1,b+2...,b+18\}$ on the plane onto $\{1,2,...18\} \times \{1,2,...18\}$ , and, as it is a function, it maps every distinct square to the same square regardless of the subgrid we are currently looking at.

This has a major implication, because if we set some fixed colouring of the grid $\{1,2,...18\} \times \{1,2,...18\}$ and define the colouring of the plane as $(x,y) \space is \space coloured \iff V(x,y) \space is \space coloured$ , every 19x19 subgrid of the plane will be isomorphic (equivalent) under $V$ to $\{1,2,...18\} \times \{1,2,...18\}$ . Consequently, every 19x19 subgrid will have the exact same number of coloured squares (in our case, 21).

Therefore, we may conclude that for any grid with dimensions $19n \times 19m$ , the minimum number of coloured squares that meets the criterion is $21nm$ . Therefore, for a $2014 \times 2014$ grid, the number is $21 \times 106^2=235956$ . The colouring of the $\{1,2,...18\} \times \{1,2,...18\}$ subgrid which minimises the number of coloured squares in the $2013 \times 2013$ subgrid of $2014 \times 2014$ is one which maximises the number of points around the perimeter. By putting all of the 21 points on the perimeter, and ensuring that one is in the corner $(19,19)$ , we maximise this number to give: $106*20+2*106-1=2331$ .

Therefore, the minimum number of coloured squares required on the $2013 \times 2013$ grid is $235956-2331=233625$ .