The natural numbers $m$ and $k$  satisfy the equality $$1001 \cdot 1002 \cdot  ... \cdot 2010 \cdot 2011 = 2^m (2k + 1)$$. Find the number $m$.
1. We start with the given product:
   \[
   1001 \cdot 1002 \cdot \ldots \cdot 2010 \cdot 2011
   \]
   This product can be expressed as:
   \[
   \frac{2011!}{1000!}
   \]

2. To find the power of 2 in the prime factorization of this product, we use Legendre's formula. Legendre's formula for the exponent of a prime \( p \) in \( n! \) is given by:
   \[
   v_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
   \]
   Here, we need to find \( v_2(2011!) \) and \( v_2(1000!) \).

3. Calculate \( v_2(2011!) \):
   \[
   v_2(2011!) = \left\lfloor \frac{2011}{2} \right\rfloor + \left\lfloor \frac{2011}{4} \right\rfloor + \left\lfloor \frac{2011}{8} \right\rfloor + \left\lfloor \frac{2011}{16} \right\rfloor + \left\lfloor \frac{2011}{32} \right\rfloor + \left\lfloor \frac{2011}{64} \right\rfloor + \left\lfloor \frac{2011}{128} \right\rfloor + \left\lfloor \frac{2011}{256} \right\rfloor + \left\lfloor \frac{2011}{512} \right\rfloor + \left\lfloor \frac{2011}{1024} \right\rfloor
   \]
   \[
   = 1005 + 502 + 251 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 2002
   \]

4. Calculate \( v_2(1000!) \):
   \[
   v_2(1000!) = \left\lfloor \frac{1000}{2} \right\rfloor + \left\lfloor \frac{1000}{4} \right\rfloor + \left\lfloor \frac{1000}{8} \right\rfloor + \left\lfloor \frac{1000}{16} \right\rfloor + \left\lfloor \frac{1000}{32} \right\rfloor + \left\lfloor \frac{1000}{64} \right\rfloor + \left\lfloor \frac{1000}{128} \right\rfloor + \left\lfloor \frac{1000}{256} \right\rfloor + \left\lfloor \frac{1000}{512} \right\rfloor
   \]
   \[
   = 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 994
   \]

5. The power of 2 in the product \( \frac{2011!}{1000!} \) is:
   \[
   v_2\left(\frac{2011!}{1000!}\right) = v_2(2011!) - v_2(1000!) = 2002 - 994 = 1008
   \]

6. Therefore, the value of \( m \) is:
   \[
   m = 1008
   \]

The final answer is \( \boxed{1008} \)