There are $2012$ backgammon checkers (stones, pieces) with one side is black and the other side is white.
These checkers are arranged into a line such that no two consequtive checkers are in same color. At each move, we are chosing two checkers. And we are turning upside down of the two checkers and all of the checkers between the two. At least how many moves are required to make all checkers same color?

$ \textbf{(A)}\ 1006 \qquad \textbf{(B)}\ 1204 \qquad \textbf{(C)}\ 1340 \qquad \textbf{(D)}\ 2011 \qquad \textbf{(E)}\ \text{None}$
1. **Initial Setup**: We have 2012 backgammon checkers arranged in a line such that no two consecutive checkers are the same color. This means the sequence alternates between black and white, e.g., BWBWBW... or WBWBWB...

2. **Grouping**: We can group the checkers into sets of 4. Specifically, we can create 503 groups where each group consists of 4 consecutive checkers. For example, the first group would be \(a_1, a_2, a_3, a_4\), the second group would be \(a_5, a_6, a_7, a_8\), and so on.

3. **Group Analysis**: Each group of 4 checkers will have the arrangement WBBWWBBW or BWBWBWBW. Let's consider the group WBBWWBBW:
   - Initial: WBBWWBBW
   - After one move: WBWBBWBW (flip the middle two checkers and the two on the ends)
   - After another move: BWBWBWBW (flip the middle two checkers and the two on the ends again)

4. **Transformation**: Each group of 4 checkers can be transformed into a sequence where all checkers are the same color in exactly 2 moves. This is because each move flips the checkers and the ones in between, effectively changing the sequence from alternating to uniform.

5. **Total Moves Calculation**: Since there are 503 groups and each group requires 2 moves to make all checkers the same color, the total number of moves required is:
   \[
   503 \times 2 = 1006
   \]

Conclusion:
\[
\boxed{1006}
\]