Consider a $(2m-1)\times(2n-1)$ rectangular region, where $m$ and $n$ are integers such that $m,n\ge 4.$ The region is to be tiled using tiles of the two types shown:
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(The dotted lines divide the tiles into $1\times 1$ squares.) The tiles may be rotated and reflected, as long as their sides are parallel to the sides of the rectangular region. They must all fit within the region, and they must cover it completely without overlapping.

What is the minimum number of tiles required to tile the region?
To solve the problem of tiling a \((2m-1) \times (2n-1)\) rectangular region using the given \(\Gamma\) and \(Z\) tiles, we need to determine the minimum number of tiles required. We will use a coloring argument and some combinatorial reasoning to achieve this.

1. **Define the Tiles and Coloring Scheme:**
   - The \(\Gamma\) tile covers 3 squares.
   - The \(Z\) tile covers 4 squares.
   - Color the \((2m-1) \times (2n-1)\) grid using four colors based on the parity of the coordinates \((a, b)\):
     - Yellow for both \(a\) and \(b\) odd.
     - Green for \(a\) even and \(b\) odd.
     - Blue for \(a\) odd and \(b\) even.
     - Red for both \(a\) and \(b\) even.

2. **Count the Number of Each Color:**
   - Yellow squares: \(mn\)
   - Green squares: \((m-1)n\)
   - Blue squares: \(m(n-1)\)
   - Red squares: \((m-1)(n-1)\)

3. **Analyze the Tiles:**
   - Each \(Z\) tile covers one square of each color.
   - Each \(\Gamma\) tile covers three squares of three different colors and omits the fourth color.

4. **Balance the Colors:**
   - To balance the colors, we need to use \(\Gamma\) tiles to account for the differences in the number of squares of each color.
   - The number of \(\Gamma\) tiles needed to balance the colors:
     - More blue \(\Gamma\) tiles than yellow: \(m\)
     - More green \(\Gamma\) tiles than yellow: \(n\)
     - More red \(\Gamma\) tiles than yellow: \(m+n-1\)

5. **Calculate the Minimum Number of \(\Gamma\) Tiles:**
   - The total number of \(\Gamma\) tiles required is at least \(m + n + (m + n - 1) = 2m + 2n - 1\).

6. **Total Number of Tiles:**
   - Let \(G\) be the number of \(\Gamma\) tiles and \(T\) be the total number of tiles.
   - The area covered by the tiles is \((2m-1)(2n-1)\).
   - The equation for the total area covered by the tiles is:
     \[
     (2m-1)(2n-1) = 3G + 4(T - G)
     \]
   - Simplifying, we get:
     \[
     (2m-1)(2n-1) = 4T - G
     \]
   - Substituting \(G \geq 2m + 2n - 1\):
     \[
     (2m-1)(2n-1) \leq 4T - (2m + 2n - 1)
     \]
   - Solving for \(T\):
     \[
     T \geq mn
     \]

7. **Conclusion:**
   - The minimum number of tiles required to tile the \((2m-1) \times (2n-1)\) rectangular region is \(mn\).

The final answer is \(\boxed{mn}\).