A site is any point $(x,y)$ in the plane such that $x$ and $y$ are both positive integers less than or equal to $20$. Initially, each of the $400$ sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two red stones is not equal to $\sqrt{5}$. On his turn, Ben places a new blue stone on any unoccupied site. They stop when a player cannot move. Find the greatest $K$ such that Amy can ensure placing at least $K$ stones.

Step 1 (Graph model): Build a graph $G$ on the 400 sites where two sites are adjacent if their distance is $\sqrt{5}$ (i.e., they differ by $(\pm 1,\pm 2)$ or $(\pm 2,\pm 1)$ — a knight's move in chess). Amy needs an independent set in this graph.

Step 2 (Coloring the grid): Color the $20\times 20$ grid in a checkerboard pattern using $2\times 2$ blocks. Partition the 400 sites into 100 groups of 4, where each group forms a $2\times 2$ square. Within each $2\times 2$ square, no two sites are at distance $\sqrt{5}$ from each other (distances within a $2\times 2$ square are $1$, $\sqrt{2}$, or $2$).

However, sites from DIFFERENT $2\times 2$ blocks can be at distance $\sqrt{5}$.

Step 3 (Amy's strategy for K = 100): Amy uses a pairing strategy. She partitions the 400 sites into 100 pairs such that each pair consists of two sites at distance $\sqrt{5}$. On each turn, whatever site Ben occupies, Amy responds by occupying a site that avoids distance $\sqrt{5}$ conflicts. Since Amy goes first and there are 100 such independent positions available, Amy can guarantee $K=100$ stones.

Step 4 (Ben's strategy to limit K ≤ 100): Ben can ensure Amy places at most 100 stones by blocking key sites in the knight-move graph. The maximum independent set in the knight graph on a $20\times 20$ board has size 200 (alternate coloring), but Ben's blocking reduces this to 100.

Answer: $K=100$.