Queenie and Horst play a game in a 20 x 20 chessboard.
In the beginning, the board is empty.
In every turn, Horst places a black knight on an empty square in such a way
that his new new knight does not attack any previous knights.
Then Queenie places a white queen on an empty square.
The game gets finished when somebody cannot move.

Find the maximal positive K such that, regardless of the strategy of Queenie,
Horst can put at least K knights on the board.

-----------------------

Let's try smaller chessboard, 2x2:
. .
. .
----
K .
. .
----
K Q
. .
----
K Q
K .
----
K Q
K Q
 --> two rounds

experiment 4x4:
. . . .
. . . .
. . . .
. . . .

How many knights can actually fit there without any blocking queens?
It cannot be more than 8 anyway -- the squares can be paired
1 3 5 7
2 4 6 8
3 1 7 5
4 2 8 6
in the knight graph.
Can we put 8 knights there?

K K K K
. . . .
. . . .
K K K K

Indeed.
But Queenie will block this.
For example, there is the 4-cycle in the knight graph:

* . . .
. . * .
. * . .
. . . *

Once one knight is put into the 4-cycle, Q can be put to te opposite
corner of the 4-cycle preventing any further K in the cycle.

Actually, the 4x4 chessboard can be split into 4 such 4-cycles:

1 2 3 4
3 4 1 2
2 1 4 3
4 3 2 1

So at most 4 knights can be placed there.
On the other hand, To each of the 4-cycles, at least one knight can be put,
so the optimal number for 4x4 is 4.

This blocking strategy can be generalized to 20x20, as 20 is divisible by 4,
so Q can get the number of Ks down to 20*20/4 = 20*5 = 100.

Can Horst actually get this number of Ks on the chessboard?
If he can get 200 Ks on the chessboard without Queenie opponent,
he can get 100 Ks with it.
Let's try smaller case again -- is it possible to place 16 Ks
on a 8x8 chessboard?

K K K K K K K K
. . . . . . . .
. . . . . . . .
K K K K K K K K
. . . . . . . .
. . . . . . . .
K K K K K K K K
. . . . . . . .
This is not good.

K K - K . . . .
K - - - . - . .
- - - . - . . .
K - . . . . . .
. . - . . . . .
. - . . . . . .
. . . . . . . .
. . . . . . . .

Ah, I am stupid

K . K . K . K .
. K . K . K . K
K . K . K . K .
. K . K . K . K
K . K . K . K .
. K . K . K . K
K . K . K . K .
. K . K . K . K

So in 20 x 20, Horst can put at least 100 knights on the
200 white squares.

The answer is 100. Problem solved!