10. (15 points) From the 2015 positive integers 1, 2, 
$\cdots, 2015$, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$.
10. Classify $1 \sim 2015$ by their remainders when divided by 50.

Let $A_{i}$ represent the set of numbers from $1 \sim 2015$ that have a remainder of $i$ when divided by 50, where $i=0,1, \cdots, 49$. Then $A_{1}, A_{2}, \cdots, A_{15}$ each contain 41 numbers, and $A_{0}, A_{16}, A_{17}, \cdots, A_{49}$ each contain 40 numbers.

According to the problem, at most one number can be selected from $A_{0}$ and $A_{25}$; and no numbers can be selected simultaneously from $A_{i}$ and $A_{50-i}$ (for $i=1,2, \cdots, 24$).

Notice that when $1 \leqslant i \leqslant 15$, $A_{i}$ contains more numbers than $A_{50-i}$.
Therefore, all numbers from $A_{1}, A_{2}, \cdots, A_{15}$ can be selected.
For $16 \leqslant i \leqslant 24$, all numbers from $A_{i}$ (or $A_{50-i}$) can be selected.
In summary, the maximum value of $k$ is
$15 \times 41 + 9 \times 40 + 1 + 1 = 977$.