Three. (50 points) Each cell of a $33 \times 33$ grid is colored with one of three colors, such that the number of cells of each color is equal. If two adjacent cells have different colors, their common edge is called a "separating edge." Determine the minimum number of separating edges.
Three, let the number of boundary lines be $L$.
First, divide the grid table into three regions as shown in Figure 3, and color them with three different colors, with all the thick lines being boundary lines.

At this point, there are 56 boundary lines, i.e., $L=56$.
Next, prove: $L \geqslant 56$.
Label the rows of the grid table from top to bottom as $A_{1}$, $A_{2}, \cdots, A_{33}$, and the columns from left to right as $B_{1}$, $B_{2}, \cdots$, $B_{33}$. The number of colors appearing in the squares of row $A_{i}$ is denoted as $n\left(A_{i}\right)$, and the number of colors appearing in the squares of column $B_{i}$ is denoted as $n\left(B_{i}\right)$. The three colors are denoted as $c_{1}$, $c_{2}$, and $c_{3}$. For a color $c_{j}$, let $n\left(c_{j}\right)$ be the sum of the number of rows and columns containing $c_{j}$-colored squares.
Define $\delta\left(A_{i}, c_{j}\right)=\left\{\begin{array}{l}1, \text { if row } A_{i} \text { contains } c_{j} \text { color; } \\ 0, \text { otherwise. }\end{array}\right.$ Similarly define $\delta\left(B_{i}, c_{j}\right)$.
Thus, $\sum_{i=1}^{33}\left(n\left(A_{i}\right)+n\left(B_{i}\right)\right)$
$=\sum_{i=1}^{33} \sum_{j=1}^{3}\left(\delta\left(A_{i}, c_{j}\right)+\delta\left(B_{i}, c_{j}\right)\right)$
$=\sum_{j=1}^{3} \sum_{i=1}^{33}\left(\delta\left(A_{i}, c_{j}\right)+\delta\left(B_{i}, c_{j}\right)\right)$
$=\sum_{j=1}^{3} n\left(c_{j}\right)$.
Since there are $\frac{1}{3} \times 33^{2}=363$ squares of color $c_{j}$, suppose there are $a$ rows and $b$ columns containing $c_{j}$-colored squares, then the $c_{j}$-colored squares must be in the intersection of these $a$ rows and $b$ columns.
Thus, $a b \geqslant 363$.
Hence, $n\left(c_{j}\right)=a+b \geqslant 2 \sqrt{a b} \geqslant 2 \sqrt{363}>38$
$\Rightarrow n\left(c_{j}\right) \geqslant 39(j=1,2,3)$.
Since there are $n\left(A_{i}\right)$ colors of squares in row $A_{i}$, there are at least $n\left(A_{i}\right)-1$ boundary lines.

Similarly, in column $B_{j}$, there are at least $n\left(B_{j}\right)-1$ boundary lines.
$$
\begin{array}{l}
\text { Then } L \geqslant \sum_{i=1}^{33}\left(n\left(A_{i}\right)-1\right)+\sum_{i=1}^{33}\left(n\left(B_{i}\right)-1\right) \\
=\sum_{i=1}^{33}\left(n\left(A_{i}\right)+n\left(B_{i}\right)\right)-66 \\
=\sum_{j=1}^{3} n\left(c_{j}\right)-66 .
\end{array}
$$

Below, we discuss two scenarios.
(1) There is a row or a column where all squares are the same color.

Without loss of generality, assume there is a row all of which are $c_{1}$ color. Then, all 33 columns of the grid table contain $c_{1}$-colored squares. Since there are 363 $c_{1}$-colored squares, at least 11 rows contain $c_{1}$-colored squares. Thus,
$$
n\left(c_{1}\right) \geqslant 11+33=44 \text {. }
$$

By equations (1), (3), and (4),
$$
\begin{array}{l}
L \geqslant n\left(c_{1}\right)+n\left(c_{2}\right)+n\left(c_{3}\right)-66 \\
\geqslant 44+39+39-66=56 .
\end{array}
$$
(2) There is no row or column where all squares are the same color.
Then, for any $1 \leqslant i \leqslant 33$,
$$
n\left(A_{i}\right) \geqslant 2, n\left(B_{i}\right) \geqslant 2 \text {. }
$$

Thus, by equation (2),
$$
\begin{array}{l}
L \geqslant \sum_{i=1}^{33}\left(n\left(A_{i}\right)+n\left(B_{i}\right)\right)-66 \\
\geqslant 33 \times 4-66=66>56 .
\end{array}
$$

In summary, the minimum number of boundary lines is 56.