We call a rectangle of the size $1 \times 2$ a domino. Rectangle of the $2 \times 3$ removing two opposite (under center of rectangle) corners we call tetramino. These figures can be rotated.

It requires to tile rectangle of size $2008  \times 2010$ by using dominoes and tetraminoes. What is the minimal number of dominoes should be used?
To solve the problem of tiling a $2008 \times 2010$ rectangle using dominoes and tetraminoes, we need to consider the properties and constraints of these shapes.

1. **Area Calculation**:
   - The area of the $2008 \times 2010$ rectangle is:
     \[
     2008 \times 2010 = 4036080
     \]
   - The area of a domino ($1 \times 2$) is:
     \[
     1 \times 2 = 2
     \]
   - The area of a tetramino (a $2 \times 3$ rectangle with two opposite corners removed) is:
     \[
     2 \times 3 - 2 = 4
     \]

2. **Tiling Constraints**:
   - Since the total area of the rectangle is $4036080$, and each domino covers $2$ units of area, and each tetramino covers $4$ units of area, we need to find a combination of these shapes that sums to $4036080$.
   - Let \( d \) be the number of dominoes and \( t \) be the number of tetraminoes. The equation for the total area covered is:
     \[
     2d + 4t = 4036080
     \]

3. **Simplifying the Equation**:
   - Divide the entire equation by 2:
     \[
     d + 2t = 2018040
     \]

4. **Minimizing the Number of Dominoes**:
   - To minimize the number of dominoes, we need to maximize the number of tetraminoes. Let’s solve for \( d \) in terms of \( t \):
     \[
     d = 2018040 - 2t
     \]
   - To ensure \( d \) is minimized, \( t \) should be maximized. The maximum value of \( t \) occurs when \( d \) is minimized, ideally \( d = 0 \):
     \[
     0 = 2018040 - 2t \implies t = 1009020
     \]
   - Substituting \( t = 1009020 \) back into the equation for \( d \):
     \[
     d = 2018040 - 2 \times 1009020 = 0
     \]

5. **Verification**:
   - If \( t = 1009020 \) and \( d = 0 \), the total area covered is:
     \[
     2 \times 0 + 4 \times 1009020 = 4036080
     \]
   - This confirms that the entire area can be covered using 1009020 tetraminoes and 0 dominoes.

Therefore, the minimal number of dominoes required is \( \boxed{0} \).