A bear is in the center of the left down corner of a $100*100$ square .we call a cycle in this grid a bear cycle if it visits each square exactly ones and gets back to the place it started.Removing a row or column with compose the bear cycle into number of pathes.Find the minimum $k$ so that in any bear cycle we can remove a row or column so that the maximum length of the remaining pathes is at most $k$.
1. **Understanding the Problem:**
   - We have a $100 \times 100$ grid.
   - A bear cycle is a path that visits each square exactly once and returns to the starting point.
   - We need to find the minimum $k$ such that removing any row or column from the grid results in the maximum length of the remaining paths being at most $k$.

2. **Initial Consideration:**
   - The total number of squares in the grid is $100 \times 100 = 10000$.
   - If we remove one row or one column, we are left with $99 \times 100 = 9900$ or $100 \times 99 = 9900$ squares.

3. **Constructing a Path:**
   - Consider a specific path that starts at the lower left corner, moves up to the top left corner, then moves right one unit, moves down 49 units, moves left one unit, moves up 49 units, and repeats this pattern until it reaches the top right corner.
   - From the top right corner, the path continues by rotating the board $180$ degrees and repeating the same pattern.

4. **Analyzing the Path:**
   - This path ensures that each square is visited exactly once and returns to the starting point.
   - If we remove the 50th row, the path will be broken into segments.
   - Each segment will have a maximum length of 5000 because the path is designed to traverse half the grid (5000 squares) before being interrupted by the removal of the row.

5. **Optimality of $k$:**
   - To show that $k = 5000$ is optimal, consider any other path that visits each square exactly once.
   - Removing any row or column will break the path into segments.
   - The maximum length of these segments cannot be less than 5000 because the path must cover 10000 squares, and removing a row or column will divide the path into two segments of approximately equal length.

6. **Conclusion:**
   - Therefore, the minimum $k$ such that removing any row or column results in the maximum length of the remaining paths being at most $k$ is $5000$.

The final answer is $\boxed{5000}$