We show two strategies, one for Horst to place at least $100$ knights, and another strategy for Queenie that prevents Horst from putting more than $100$ knights on the board.

A strategy for Horst: Put knights only on black squares, until all black squares get occupied.

Colour the squares of the board black and white in the usual way, such that the white and black squares alternate, and let Horst put his knights on black squares as long as it is possible. Two knights on squares of the same colour never attack each other. The number of black squares is $20^{2} / 2 = 200$. The two players occupy the squares in turn, so Horst will surely find empty black squares in his first $100$ steps.

A strategy for Queenie: Group the squares into cycles of length $4$, and after each step of Horst, occupy the opposite square in the same cycle.

Consider the squares of the board as vertices of a graph; let two squares be connected if two knights on those squares would attack each other. Notice that in a $4 \times 4$ board the squares can be grouped into $4$ cycles of length $4$, as shown in Figure 1. Divide the board into parts of size $4 \times 4$, and perform the same grouping in every part; this way we arrange the $400$ squares of the board into $100$ cycles (Figure 2).

Figure 1

Figure 2

Figure 3

The strategy of Queenie can be as follows: Whenever Horst puts a new knight to a certain square $A$, which is part of some cycle $A-B-C-D-A$, let Queenie put her queen on the opposite square $C$ in that cycle (Figure 3). From this point, Horst cannot put any knight on $A$ or $C$ because those squares are already occupied, neither on $B$ or $D$ because those squares are attacked by the knight standing on $A$. Hence, Horst can put at most one knight on each cycle, that is at most $100$ knights in total.