Starting from $0$, at each step we take $1$ more or $2$ times of the previous number. Which one below can be get in a less number of steps?

$ \textbf{(A)}\ 2011
\qquad\textbf{(B)}\ 2010
\qquad\textbf{(C)}\ 2009
\qquad\textbf{(D)}\ 2008
\qquad\textbf{(E)}\ 2007
$
1. We need to determine the number of steps required to reach each of the given numbers starting from 0, using the operations "add 1" or "double the previous number".
2. The number of steps to reach a number \( n \) can be interpreted as the total number of bits in the binary representation of \( n \), minus one, plus the number of ones in the binary representation of \( n \). This is because each bit represents a doubling operation, and each '1' in the binary representation represents an addition operation.

Let's convert each number to its binary form and count the number of bits and the number of ones:

- \( 2007_{10} = 11111010111_2 \)
  - Number of bits: 11
  - Number of ones: 8
  - Total steps: \( 11 - 1 + 8 = 18 \)

- \( 2008_{10} = 11111011000_2 \)
  - Number of bits: 12
  - Number of ones: 5
  - Total steps: \( 12 - 1 + 5 = 16 \)

- \( 2009_{10} = 11111011001_2 \)
  - Number of bits: 12
  - Number of ones: 6
  - Total steps: \( 12 - 1 + 6 = 17 \)

- \( 2010_{10} = 11111011010_2 \)
  - Number of bits: 12
  - Number of ones: 6
  - Total steps: \( 12 - 1 + 6 = 17 \)

- \( 2011_{10} = 11111011011_2 \)
  - Number of bits: 12
  - Number of ones: 7
  - Total steps: \( 12 - 1 + 7 = 18 \)

3. Comparing the total steps for each number:
   - 2007: 18 steps
   - 2008: 16 steps
   - 2009: 17 steps
   - 2010: 17 steps
   - 2011: 18 steps

4. The number that can be reached in the least number of steps is 2008.

The final answer is \(\boxed{2008}\).