Four, (50 points) Color each cell of a $5 \times 5$ grid with one of five colors, such that the number of cells of each color is the same. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.

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As shown in Figure 4, a $5 \times 5$ grid is divided into five parts, each colored with one of 5 colors, and at this point, there are 16 boundary edges.
The following proves that the number of boundary edges is at least 16.
First, let the 5 colors be $1, 2, \cdots, 5$, and let the number of rows and columns occupied by the cells of the $i$-th color be denoted as $a_{i}$ and $b_{i}$, respectively, and the length of the boundary be denoted as $c_{i}$.

For $i \in \{1, 2, \cdots, 5\}$, the $i$-th color is exactly applied to 5 cells, and it occupies $a_{i}$ rows and $b_{i}$ columns, thus
$$
a_{i} b_{i} \geqslant 5 \text {. }
$$

Since for each row among the $a_{i}$ rows occupied by the $i$-th color, the left and right boundaries of the $i$-th color cells in this row must be the boundaries of all the $i$-th color cells, and for each column among the $b_{i}$ columns occupied by the $i$-th color, the upper and lower boundaries of the $i$-th color cells in this column must also be the boundaries of all the $i$-th color cells, therefore,
$$
c_{i} \geqslant 2\left(a_{i}+b_{i}\right) \text {. }
$$

Let the number of boundary edges be $f$. Since each boundary edge must be the boundary between two colors, and the boundary of the $i$-th color cells, excluding the boundary of the $5 \times 5$ grid, must be a boundary edge, thus,
$$
f=\frac{1}{2}\left(\sum_{i=1}^{5} c_{i}-5 \times 4\right) .
$$

From $a_{i} b_{i} \geqslant 5$, we know $a_{i}+b_{i} \geqslant 2 \sqrt{a_{i} b_{i}} \geqslant 2 \sqrt{5}$.
Since $a_{i}, b_{i} \in \mathbf{Z}$, it follows that,
$$
a_{i}+b_{i} \geqslant 5 \Rightarrow c_{i} \geqslant 10 \text {, }
$$

and equality holds if and only if the $i$-th color cells are arranged as shown in Figure 5, denoted as Figure A.
$$
\begin{array}{l}
\text { Hence } f \geqslant \frac{1}{2}(5 \times 10-5 \times 4) \\
=15,
\end{array}
$$

and the equality condition is that there are 5 Figure A's.

Notice that 5 $2 \times 2$ grids cannot be arranged without overlapping in a $5 \times 5$ grid.
Thus, the equality in (1) does not hold.
Therefore, $f \geqslant 16$.