Three, (50 points) Number the 64 squares of an $8 \times 8$ grid as $1,2, \cdots, 64$, such that for all $1 \leqslant i \leqslant 63$, the squares numbered $i$ and $i+1$ share a common edge. Find the maximum possible sum of the numbers on the eight squares along the main diagonal.
Three, for the sake of convenience, the cell numbered $a$ is simply referred to as cell $a$. If the grid is colored alternately in black and white like a chessboard, then because cell $a$ and cell $a+1$ share a common edge, cell $a$ and cell $a+1$ are of different colors. Therefore, the eight cells on the main diagonal are of the same color, meaning these eight cells have numbers of the same parity. Let the smallest number among these eight numbers be $m$.

Since there are 28 cells on each side of the main diagonal, and cell $a$ and cell $a+1$ share a common edge, and $m$ is the smallest number among the eight cells on the main diagonal, the numbers $1, 2, \cdots, m-1$ must be located on the same side of the main diagonal, so $m \leqslant 29$.

When $m$ is odd, the eight numbers on the main diagonal are all odd, and their maximum possible sum is
$$
29+51+53+55+57+59+61+63=428 \text {. }
$$

When $m$ is even, note that among the 28 cells on the same side of the main diagonal, 16 cells are of one color, and 12 cells are of the other color. Consider the path formed by cells of alternating colors. Before it intersects with the main diagonal, it can pass through at most $13+12=25$ steps, so $m \leqslant 26$. Also, the eight numbers on the main diagonal are all even, and their maximum possible sum is
$$
26+52+54+56+58+60+62+64=432,
$$

as shown in Figure 3, this value can be achieved.
Since $432>428$, the maximum value sought is 432.