Example 3. In an arithmetic sequence with the first term $a_{1}>0$, the sum of the first 9 terms is equal to the sum of the first 17 terms. How many terms have the maximum sum?
This is a common problem, usually solved by completing the square or solving a system of inequalities, which can be cumbersome. Let's first analyze the implicit conditions in the problem: (1) It is a decreasing sequence (2) $a_{10}+a_{11}+a_{12}+a_{13}+a_{14}+a_{1 \mathrm{~s}}+a_{16}+a_{17}=0$. By the properties of an arithmetic sequence, we know that $a_{10}+a_{17}=a_{11}+a_{16}=a_{12}+a_{15}=a_{1 \mathrm{~s}}+a_{14}$. Therefore, the equation in (2) can be simplified to $4\left(a_{13}+a_{14}\right)=0$. Observing the structure of this equation, and given that $a_{1}>0$ and $\left\{a_{n}\right\}$ is decreasing, we can deduce that $a_{13}>0$ and $a_{14}<0$. Therefore, the sum of the first 13 terms is the maximum.

In the process of solving this problem, we successively identified and utilized three special conditions that emerged during the solution, simplifying the steps.