First assume the opposite: it is possible to put all 100 queens on the chessboard so that in one 50x50 subboard there are no queens.
The chessboard can roughly be drawn as follows, where no queens can go in the X'ed part.
XXXXX----- XXXXX----- XXXXX--A-- XXXXX----- XXXXX----- ---------- ---------- --B----C-- ---------- ----------
Clearly, in each row there must be exactly one queen. If there are two or more, those queens would attack each other; if there are none, by the pigeonhole principle some other row will have two or more queens. In particular, in the top 50 rows there must be 50 queens, and since they can't go in the X'ed subboard they must all go in the A subboard.
By the same reasoning with columns, subboard B must also have 50 queens so each column will have a queen in it.
Therefore, there are no queens left for subboard C; it must also be empty.
50 +---+ | | XXXXX///// XXXXX/////-+ XXXXX//A// | 49 XXXXX///// | XXXXX/////-+ /////XXXXX /////XXXXX //B//XXCXX /////XXXXX /////XXXXX
Now consider the diagonals that run from A to B. There are only 99 of them that go through A and B. The queens can't be on any other diagonals, because they are limited to the subboards A and B. By the pigeonhole principle at least one of the diagonals must have two or more queens on it; those queens attack each other. Contradiction. Therefore it is impossible to put all 100 queens on the board this way. |