The beaver is chess piece that move to $2$ cells by horizontal or vertical.  Every cell of $100 \times 100$ chessboard colored in some color,such that we can not get from one cell to another with same color with one move of beaver or knight. What minimal color do we need?
To solve this problem, we need to determine the minimum number of colors required to color a $100 \times 100$ chessboard such that no two cells of the same color are reachable by a single move of a beaver or a knight.

1. **Understanding the Moves:**
   - A beaver moves exactly 2 cells horizontally or vertically.
   - A knight in chess moves in an "L" shape: 2 cells in one direction and 1 cell perpendicular to that direction.

2. **Coloring with 4 Colors:**
   - We can use a 4-color pattern to ensure that no two cells of the same color are reachable by a beaver or a knight move.
   - The pattern can be repeated every 4 rows and 4 columns. Here is an example of such a pattern:

\[
\begin{array}{cccccccc}
1 & 1 & 2 & 2 & 1 & 1 & 2 & 2 \\
1 & 1 & 2 & 2 & 1 & 1 & 2 & 2 \\
3 & 3 & 4 & 4 & 3 & 3 & 4 & 4 \\
3 & 3 & 4 & 4 & 3 & 3 & 4 & 4 \\
1 & 1 & 2 & 2 & 1 & 1 & 2 & 2 \\
1 & 1 & 2 & 2 & 1 & 1 & 2 & 2 \\
3 & 3 & 4 & 4 & 3 & 3 & 4 & 4 \\
3 & 3 & 4 & 4 & 3 & 3 & 4 & 4 \\
\end{array}
\]

3. **Verification:**
   - For the beaver move: Any cell of color 1 is surrounded by cells of colors 2, 3, and 4, and no two cells of the same color are 2 cells apart horizontally or vertically.
   - For the knight move: Any cell of color 1 is surrounded by cells of colors 2, 3, and 4 in the knight's move positions.

4. **Proof of Impossibility with 3 Colors:**
   - Assume we try to color the board with 3 colors: A, B, and C.
   - Consider a central cell \( L \) colored A. The cells reachable by a beaver or knight move from \( L \) must be colored B or C.
   - This leads to a contradiction as shown in the problem's diagram. Specifically, there will be a cell that cannot be colored without violating the condition that no two cells of the same color are reachable by a beaver or knight move.

\[
\begin{array}{|c|c|c|c|c|}
\hline 
 & P &  & Q & \\
\hline 
 &  & X &  & \\
\hline 
R &  & L &  & S \\
\hline 
 &  & Y &  & \\
\hline 
 & P' &  & Q' & \\
\hline 
\end{array}
\]

- If \( L = A \), \( P = B \), \( Q = C \), and so on, we eventually find that there is no valid color for some cells, leading to a contradiction.

Conclusion:
Thus, it is impossible to color the board with only 3 colors while satisfying the given conditions. Therefore, the minimum number of colors required is 4.

The final answer is \( \boxed{4} \)