There are $ 6n \plus{} 4$ mathematicians participating in a conference which includes $ 2n \plus{} 1$ meetings. Each meeting has one round table that suits for $ 4$ people and $ n$ round tables that each table suits for $ 6$ people. We have known that two arbitrary people sit next to or have opposite places doesn't exceed one time. 
1. Determine whether or not there is the case $ n \equal{} 1$.
2. Determine whether or not there is the case $ n > 1$.
1. **Determine whether or not there is the case \( n = 1 \).**

   For \( n = 1 \):
   - The total number of mathematicians is \( 6n + 4 = 6(1) + 4 = 10 \).
   - The number of meetings is \( 2n + 1 = 2(1) + 1 = 3 \).
   - Each meeting has one 4-person table and \( n = 1 \) 6-person table.

   We need to check if we can arrange 10 people in 3 meetings such that no two people sit next to or opposite each other more than once.

   Let's divide the 10 people into 4 teams: one team of one member (let's call this person \( A \)), and 3 teams of three members each.

   We can arrange a tournament of 3 rounds such that any two teams meet exactly once. In each round, one team meets \( A \) at the 4-person table, and the other 2 teams meet at the 6-person table.

   - **Round 1:**
     - 4-person table: \( A \) and Team 1 (3 members)
     - 6-person table: Team 2 (3 members) and Team 3 (3 members)

   - **Round 2:**
     - 4-person table: \( A \) and Team 2 (3 members)
     - 6-person table: Team 1 (3 members) and Team 3 (3 members)

   - **Round 3:**
     - 4-person table: \( A \) and Team 3 (3 members)
     - 6-person table: Team 1 (3 members) and Team 2 (3 members)

   In each round, the arrangement ensures that any two people sit next to or opposite each other exactly once. Therefore, there is a valid arrangement for \( n = 1 \).

2. **Determine whether or not there is the case \( n > 1 \).**

   For \( n > 1 \):
   - The total number of mathematicians is \( 6n + 4 \).
   - The number of meetings is \( 2n + 1 \).
   - Each meeting has one 4-person table and \( n \) 6-person tables.

   We need to check if we can arrange \( 6n + 4 \) people in \( 2n + 1 \) meetings such that no two people sit next to or opposite each other more than once.

   Let's divide the \( 6n + 4 \) people into \( 2n + 2 \) teams: one team of one member (let's call this person \( A \)), and \( 2n + 1 \) teams of three members each.

   We can arrange a tournament of \( 2n + 1 \) rounds such that any two teams meet exactly once. In each round, one team meets \( A \) at the 4-person table, and the other \( 2n \) teams meet at the \( n \) 6-person tables.

   - **Round 1:**
     - 4-person table: \( A \) and Team 1 (3 members)
     - \( n \) 6-person tables: The remaining \( 2n \) teams meet in pairs.

   - **Round 2:**
     - 4-person table: \( A \) and Team 2 (3 members)
     - \( n \) 6-person tables: The remaining \( 2n \) teams meet in pairs.

   - Continue this pattern for \( 2n + 1 \) rounds.

   In each round, the arrangement ensures that any two people sit next to or opposite each other exactly once. Therefore, there is a valid arrangement for \( n > 1 \).

The final answer is "yes" for both \( \boxed{ n = 1 } \) and \( n > 1 \).