10. Select 1004 numbers from $1,2, \cdots, 2008$ such that their total sum is 1009000, and the sum of any two of these 1004 numbers is not equal to 2009. Then the sum of the squares of these 1004 numbers is $\qquad$
Reference formula:
$$
1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1) \text {. }
$$
10.1351373940.

Divide $1,2, \cdots, 2008$ into 1004 groups:
$$
\{1,2008\},\{2,2007\}, \cdots,\{1004,1005\} \text {. }
$$

By the problem's setup, exactly one number is taken from each group. Replace 2, 4, $\cdots, 2008$ with 1004, 1006, 1008, 1010 by their counterparts in the same group, 1005, 1003, 1001, 999, respectively, and keep the other numbers unchanged. This results in the 1004 numbers that meet the problem's requirements.
$$
\begin{array}{l} 
2+ 4+\cdots+2008-(1004+1006+1008+ \\
1010)+(1005+1003+1001+999) \\
= 1009020-(-1+3+7+11) \\
= 1009000, \\
2^{2}+4^{2}+\cdots+2008^{2}- \\
\left(1004^{2}+1006^{2}+1008^{2}+1010^{2}\right)+ \\
\left(1005^{2}+1003^{2}+1001^{2}+999^{2}\right) \\
= 4\left(1^{2}+2^{2}+\cdots+1004^{2}\right)- \\
2009(-1+3+7+11) \\
= \frac{2}{3} \times 1004 \times 1005 \times 2009-2009 \times 20 \\
= 2008 \times 335 \times 2009-40180 \\
= 1351373940 .
\end{array}
$$

The answer is independent of the selection method.
(Xia Xingguo provided)