10. (6 points) From the natural numbers 1, 2, 3, 4, ..., 2013, the maximum number of numbers that can be taken is $\qquad$ 1008, so that the difference between any two of these numbers is not equal to 9.
【Solution】Solution: Group the natural numbers $1,2,3,4, \cdots 2013$ into sets of 9; i.e., $(1,2,3,4,5,6$, $7,8,9)$;
$$
\begin{array}{l}
(10,11,12,13,14,15,16,17,18) ; \cdots( \\
1999,2000,2001,2002,2003,2004,2005,2006,2007) ;
\end{array}
$$

The remaining 6 numbers form 1 group; select the odd-numbered groups, which is $223 \div 2=111$ (groups) $\cdots 1$ (group)
plus any two numbers from the 6 remaining numbers 2008, 2009, ... 2013 whose difference does not equal 9, but the last 6 remaining numbers 2008, 2009, ... 2013 cannot be chosen,

otherwise, with the last odd-numbered group 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, there will be a situation where the difference between any two numbers equals 9.

The maximum number of numbers that can be chosen, ensuring that the difference between any two numbers is not equal to 9, is $111 \times 9+9=1008$. Therefore, the answer is: 1008.