Example 9. Primary School One and Primary School Two have the same number of students participating in the Golden Cup Competition. The schools use cars to transport the students to the examination site. Primary School One uses cars that can seat 15 people each; Primary School Two uses cars that can seat 13 people each. As a result, Primary School Two has to send one more car than Primary School One. Later, each school adds one more student to the competition, and this way, the number of cars needed by both schools becomes the same. Finally, it is decided that each school will add one more student to the competition, and once again, Primary School Two has to send one more car than Primary School One. How many students from both schools will participate in the competition in the end?
Let's assume that at the beginning, both schools had $x$ participants. According to the problem, the first school sent $\frac{x}{15}=a \cdots 0$ ($a$ cars),

the second school sent $\frac{x}{13}=a \cdots 12(a+1$ cars $)$; the second time, the first school sent $\frac{x+1}{15}=a \cdots 1(a+1$ cars $)$,

the second school sent $\frac{x+1}{13}=a+1 \cdots 0(a+1$ cars $)$; the third time, the first school sent $\frac{x+2}{15}=a \cdots 2(a+1$ cars $)$,

the second school sent $\frac{x+2}{13}=a+1 \cdots 1(a+2$ cars). Therefore, $x+1$ is not a multiple of 13.

Thus, $x=90$.

Answer: In total, both schools had 184 participants.