all $n$

We will prove that any positive integer $n \geq 3 $ satisfies the condition. In the competition, each student is evaluated based on three criteria: juggling ability, singing ability, and math skills. A student who ranks higher in each criterion can defeat a student who ranks lower in that criterion. Considering the juggling ability and singing ability, there is no student who can defeat all students in both criteria. Suppose there are are $n$ students, with criteria pairs being $(a_1, b_n), (a_2, b_{n-1}), \ldots, (a_n, b_1)$. Suppose $ a_1 \geq a_2 \geq \cdots \geq a_n$ and $b_1 \geq b_2 \geq \cdots \geq b_n$. For every $1 \leq i \leq n $, suppose the student with the best maths skills and the juggling ability not weaker than $a_i$ or the singing ability not weaker than $b_{n-i-1}$ is $P_i$. We proceed by classification discussion: if the juggling ability of $P_i$ is not weaker than $a_i$, then $P_i$ can defeat all students whose singing ability is not weaker than $b_{n-i-1}$. Similarly, if $P_i$'s singing ability is not weaker than $ b_{n-i-1}$, then $P_i$ can defeat all students whose juggling ability is not weaker than $a_i$. For each $P_i$, suppose the types are the same, and and take $1 \leq i_0 < n $. WLOG, suppose $P_{i_0}$'s juggling ability is not weaker than $a_{i_0}$ and $P_{i_0+1} $'s singing ability is not weaker than $ b_{n-i_0-2}$. Then, from the above analysis, $ P_{i_0+1}$ can defeat all students whose juggling ability is not weaker than $a_{i_0+1}$ and $P_{i_0}$ defeats all students whose singing ability is not weaker than $b_{n-i_0-1}$. Therefore, among the remaining students, one must satisfy that the juggling ability is not stronger than $a_{i_0+1}$ and the singing ability is not stronger than $b_{n-i_0-1}$. Taking $(a_{i_0},b_{n+1-i_0})$ satisfies the condition as required.