At a party, there are $2011$ people with a glass of fruit juice each sitting around a circular table. Once a second, they clink glasses obeying the following two rules:
(a) They do not clink glasses crosswise.
(b) At each point of time, everyone can clink glasses with at most one other person.
How many seconds pass at least until everyone clinked glasses with everybody else?

[i](Swiss Mathematical Olympiad 2011, Final round, problem 1)[/i]
1. **Understanding the Problem:**
   We have 2011 people sitting in a circle, each with a glass of fruit juice. They need to clink glasses with every other person exactly once, following two rules:
   - They do not clink glasses crosswise.
   - At each point in time, everyone can clink glasses with at most one other person.

2. **Initial Observations:**
   - Each person needs to clink glasses with 2010 other people.
   - At each second, each person can clink glasses with one other person, meaning there are 1005 pairs clinking glasses simultaneously (since 2011 is odd, one person will be left out each second).

3. **Pairing Strategy:**
   - We can pair people such that no two pairs cross each other. For example, in the first second, we can pair:
     \[
     (A_1, A_2), (A_3, A_4), \ldots, (A_{2010}, A_{2011})
     \]
   - In the next second, we can shift the pairs by one position:
     \[
     (A_1, A_3), (A_2, A_4), \ldots, (A_{2009}, A_{2011}), (A_{2010}, A_1)
     \]
   - Continue this process, shifting the pairs each second.

4. **Counting the Number of Seconds:**
   - Each person needs to clink glasses with 2010 other people.
   - In each second, each person clinks glasses with one new person.
   - Therefore, it will take 2010 seconds for each person to clink glasses with every other person.

5. **Verification:**
   - We need to ensure that the pairing strategy covers all possible pairs without repetition.
   - By shifting the pairs each second, we ensure that each person clinks glasses with a new person every second.
   - Since there are 2010 other people to clink glasses with, and each second introduces a new pair, it will indeed take 2010 seconds for everyone to clink glasses with everyone else.

The final answer is $\boxed{2010}$