2. $K=100$.

First, Player A has a strategy to ensure placing at least 100 red stones.

Divide all positions into odd and even categories. If $2 \nmid (x+y)$, then the position $(x, y)$ is called an odd position; otherwise, it is called an even position.

Since the distance between any two odd positions is not equal to $\sqrt{5}$, and there are 200 odd positions, Player A can place a red stone in an empty odd position during each of the first 100 turns, thus ensuring the placement of at least 100 red stones.

Second, Player B has a strategy to prevent Player A from placing 101 red stones.

Consider a $4 \times 4$ grid, which can be divided into four groups, as shown in Figure 2, with positions marked with the same letter forming one group.

In the same group of four positions, connect the two points that are $\sqrt{5}$ units apart, forming a parallelogram. Player B adopts the following strategy: First, divide the 400 positions into 25 $4 \times 4$ grids, and in each grid, divide them into four groups of four points each, as shown in Figure 2. Connect the two points that are $\sqrt{5}$ units apart in each group, forming 100 parallelograms. Whenever Player A chooses a vertex in a parallelogram to place a red stone $P$, Player B places a blue stone at the opposite vertex of the same parallelogram. This prevents Player A from placing a red stone at the two adjacent vertices. Thus, Player A can place at most one red stone in each parallelogram.
Therefore, Player A cannot place 101 red stones.
In conclusion, the required $K=100$.
2. Define a "position" as a point $(x, y)$ on a rectangular coordinate plane, where $x$ and $y$ are positive integers not exceeding 20. Initially, all 400 positions are empty. Two players, A and B, take turns placing stones, with A starting first. Each time it is A's turn, he places a new red stone on an empty position, with the requirement that the distance between any two red stones is not equal to $\sqrt{5}$. Each time it is B's turn, he places a new blue stone on any empty position (the distance between the blue stone and other stones can be any value). This continues until one of the players can no longer place a stone. Determine the largest integer $K$ such that, regardless of how B places the blue stones, A can always ensure placing at least $K$ red stones.
（IMO Problem 4 this year）