9. (40 points) The numbers $s_{1}, s_{2}, \ldots, s_{1008}$ are such that their sum is $2016^{2}$. It is known that

$$
\frac{s_{1}}{s_{1}+1}=\frac{s_{2}}{s_{2}+3}=\frac{s_{3}}{s_{3}+5}=\ldots=\frac{s_{1008}}{s_{1008}+2015}
$$

Find $s_{17}$.
Answer: 132.

Solution. Note that none of the $s_{i}$ is equal to zero (otherwise, all the fractions $\frac{s_{i}}{s_{i}+2 i-1}$ would be equal to zero, and, consequently, all $s_{i}$ would have to be equal to zero, which contradicts the fact that their sum is $2016^{2}$). Therefore, the original condition is equivalent to the condition

$$
\frac{s_{1}+1}{s_{1}}=\frac{s_{2}+3}{s_{2}}=\frac{s_{3}+5}{s_{3}}=\ldots=\frac{s_{1008}+2015}{s_{1008}} \Leftrightarrow \frac{1}{s_{1}}=\frac{3}{s_{2}}=\frac{5}{s_{3}}=\ldots=\frac{2015}{s_{1008}}
$$

From the first equality, express $s_{2}$ in terms of $s_{1}: s_{2}=3 s_{1}$; then, similarly, $s_{3}$ in terms of $s_{1}$: $s_{3}=5 s_{1}; \ldots s_{1008}=2015 s_{1}$. Now we have

$$
\sum_{i=1}^{1008} s_{i}=\sum_{i=1}^{1008}(2 i-1) s_{1}=s_{1} \cdot \frac{(1+2015) \cdot 1008}{2}=1008^{2} s_{1}=2016^{2}
$$

From this, $s_{1}=4$ and, consequently, $s_{17}=(2 \cdot 17-1) \cdot 4=132$.