First,consider a random black square,and let the side of each square be 1.Now,construct a Big square such that the centre of it is the same as the centre of the black square,and the.side of the Big square be 3.Now,all of the white squares must be in that Big square. Lemma: If we have any finite number of squares which can be obtained from each other by translation,and every two have a common point,then all of them have a common point. Proof: We consider a coordinate system with X-asis parallel to one,and Y-asis parallel to the other side of the square.Now,we can easyly obtain that there exist a square with side 2 such that all the squares are in this square.Now,all of the squares pass through the center of this square,so the proof of the lemma is finished. The main proof: Now,we will have 2 cases: Case 1) There exist two black squares that don't intersect and two white squares that don't intersect. Now,if two black(white) squares don't intersect,white(black) squares are in the intersection area of the two Big squares for this two picked squares,and the inersecrion area is a rectangle whose 1 side is less or equal to 3,and one side is less than 2.Now,for every rectangle with this property we can pick two points inside it,such that every square side 1 with parallel sides that it is in the rectangle passes through at least one point.So,we consider the even case,the odd is the same.We have for sure n/2 black squares thart intersect each other,and at least n/2 white squares that intesect each other,so applying this lemma we are finished. Case 2) All black squares intersect each other,or all white squares intersect each other. Applying the lemma we directly have the desired result.