The ninety-seventh question: On a $999 \times 999$ grid board, there is a lame bird moving according to the following rules: from the center of any cell to the center of any adjacent cell (cells that share a common edge), each move must turn, meaning that any two consecutive moves must be perpendicular. A non-intersecting path of the lame bird refers to the centers of the cells it passes through, all of which are distinct. If the lame bird can, after reaching the last cell center of this path, directly move to the first cell center of this path according to the above rules, then this non-intersecting path is called "closed." How many cells does the longest closed, non-intersecting path of the lame bird pass through?
The 97th question:
Answer: The longest closed and non-intersecting path of the lame bird can pass through exactly $4\left(499^{2}-1\right)$ squares.
First, we prove: this number is the upper bound of the number of squares the lame bird can pass through.
Let the center of the square in the $\mathrm{i}$-th row and $\mathrm{j}$-th column be denoted as $(\mathrm{i}, \mathrm{j})$. According to the following method, each square is colored with one of the four colors $A, B, C, D$:
If $(\mathrm{i}, \mathrm{j}) \equiv(0,0)(\bmod 2)$, then the square centered at $(\mathrm{i}, \mathrm{j})$ is colored $\mathrm{A}$;
If $(\mathrm{i}, \mathrm{j}) \equiv(0,1)(\bmod 2)$, then the square centered at $(\mathrm{i}, \mathrm{j})$ is colored $B$;
If $(\mathrm{i}, \mathrm{j}) \equiv(1,0)(\bmod 2)$, then the square centered at $(\mathrm{i}, \mathrm{j})$ is colored $\mathrm{C}$;
If $(\mathrm{i}, \mathrm{j}) \equiv(1,1)(\bmod 2)$, then the square centered at $(\mathrm{i}, \mathrm{j})$ is colored $\mathrm{D}$.
The lame bird can only move from a square of color $\mathrm{A}$ to a square of color $\mathrm{B}$ or $\mathrm{C}$.
In the first case, the sequence of colors the lame bird passes through is: $A \rightarrow B \rightarrow D \rightarrow C \rightarrow A \rightarrow B \rightarrow D \rightarrow$ $\mathrm{C} \rightarrow \mathrm{A} \rightarrow \ldots$;

In the second case, the sequence of colors the lame bird passes through is: $\mathrm{A} \rightarrow \mathrm{C} \rightarrow \mathrm{D} \rightarrow \mathrm{B} \rightarrow \mathrm{A} \rightarrow \mathrm{C} \rightarrow \mathrm{D} \rightarrow$ $\mathrm{B} \rightarrow \mathrm{A} \rightarrow \ldots$.
By symmetry, we only discuss the first case.
Since the path is closed and the number of squares of each color on this path is the same, we only need to consider the number of squares of color $\mathrm{A}$ on this path.

Notice that there are $499^{2}$ squares of color $\mathrm{A}$. We will prove below that the lame bird cannot pass through all squares of color $\mathrm{A}$. Under this conclusion, the path that satisfies the condition can contain at most $4\left(499^{2}-1\right)$ squares.

Assume there is a path that passes through all squares of color $\mathrm{A}$, and color the squares of color $\mathrm{A}$ in a checkerboard pattern, i.e., any two squares of color $\mathrm{A}$ with a center distance of 2 are colored differently. Since the number of squares of color $\mathrm{A}$ is odd, the lame bird cannot always alternate between black and white along this path. Therefore, there exist two squares of the same color that were initially of color $\mathrm{A}$, such that one square can reach the other in 4 steps. Let's assume these two squares are $(\mathrm{a}, \mathrm{b})$ and $(\mathrm{a}+2, \mathrm{~b}+2)$.

As shown in Figure (1), the path from $(a, b)$ to $(a+2, b+2)$ is unique, which can be: $(a, b) \rightarrow$ $(a, b+1) \rightarrow(a+1, b+1) \rightarrow(a+1, b+2) \rightarrow(a+2, b+2)$.
Therefore, the color of $(a, b+1)$ is $B$.
Next, consider the square of color $\mathrm{A}$ at $(\mathrm{a}, \mathrm{b}+2)$.
In the first case, the lame bird must pass through $\mathrm{B}$ after passing through $\mathrm{A}$, so the path through $(\mathrm{a}, \mathrm{b}+2)$ must be: $(a-1, b+2) \rightarrow(a, b+2) \rightarrow(a, b+3)$. Thus, the path connecting these two segments must be the path connecting $(a, b+3)$ and $(a, b)$, and the path connecting $(a+2, b+2)$ and $(a-1, b+2)$. However, the centers of these four unit squares form the opposite vertices of a convex quadrilateral, and the path is outside this quadrilateral, so these two paths must intersect. Therefore, we can conclude:

The path connecting $(\mathrm{a}, \mathrm{b}+3)$ and $(\mathrm{a}, \mathrm{b})$ and the line segment connecting these two points form a closed loop, making $(\mathrm{a}-1, \mathrm{~b}+2)$ and $(a+2, b+2)$ one inside and one outside this loop. Thus, the path connecting these two points must cross the previous closed loop, meaning the intersection point is passed twice by the lame bird. Contradiction!
Therefore, the number of squares the lame bird can pass through is at most $4\left(499^{2}-1\right)$.
On the other hand, we provide an example that passes through exactly $4\left(499^{2}-1\right)$ squares.
The following Figure (2) shows that for $n \equiv 3(\bmod 4)$, an $n \times n$ chessboard can use a recursive method to construct a path that satisfies the condition and exactly does not pass through one square of color $A$ (Figure (2) shows a $15 \times 15$ chessboard, with the center marked with a black dot being the square of color $A$ that is not passed through).
In summary, the longest closed and non-intersecting path of the lame bird can pass through exactly $4\left(499^{2}-1\right)$ squares.