10. (6 points) The prize for the mathematics competition team award is 10000 copies of mathematics extracurricular reading materials. The prizes are distributed to the schools of the top five representative teams. The team with a higher ranking receives more copies, and the number of copies awarded for each ranking is a multiple of 100. If the number of copies received by the first place is the sum of the copies received by the second and third places, and the number of copies received by the second place is the sum of the copies received by the fourth and fifth places. Then, the maximum number of copies the third place can receive is $\qquad$ copies.
【Solution】Solution: Let the third place receive $x$ books, then the second place receives at least $(x+100)$ books, and the first place receives at least $(2 x+100)$ books,
$$
\begin{array}{l}
2 x+100+x+100+x+x+100 \leqslant 10000, \\
5 x+300 \leqslant 10000, \\
5 x \leqslant 9700, \\
x \leqslant 1940,
\end{array}
$$

Also, because the number of books the first place receives is the sum of the number of books the second and third places receive, 1900 does not meet the requirements,

Therefore, when using 1800 to restore, the sum of the first to fifth places has no solution, so the third place can receive at most 1700 books. Answer: The third place can receive at most 1700 books, so the answer is: 1700.