The $n$ contestant of EGMO are named $C_1, C_2, \cdots C_n$. After the competition, they queue in front of the restaurant according to the following rules.
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[*]The Jury chooses the initial order of the contestants in the queue.
[*]Every minute, the Jury chooses an integer $i$ with $1 \leq i \leq n$.
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[*]If contestant $C_i$ has at least $i$ other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly $i$ positions.
[*]If contestant $C_i$ has fewer than $i$ other contestants in front of her, the restaurant opens and process ends.
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[*]Prove that the process cannot continue indefinitely, regardless of the Jury’s choices.
[*]Determine for every $n$ the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves.
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### Part (a): Prove that the process cannot continue indefinitely

1. **Initial Setup and Definitions**:
   - Let \( C_1, C_2, \ldots, C_n \) be the contestants.
   - Define \( f(i) \) as the number of moves contestant \( C_i \) can make.

2. **Base Case**:
   - Clearly, \( f(n) = 0 \) because \( C_n \) will always have fewer than \( n \) people in front of her.

3. **Recursive Step**:
   - If \( C_k \) can make a move, there are at least \( k \) contestants in front of her.
   - One of these contestants will be \( C_\ell \) with \( \ell > k \).
   - \( C_k \) can skip over \( C_\ell \) at most \( 1 + f(\ell) \) times because every time \( C_k \) skips over \( C_\ell \), \( C_\ell \) must make a move to be skipped over again.

4. **Inequality for Moves**:
   - We can write:
     \[
     f(k) \leq \sum_{\ell > k} (1 + f(\ell))
     \]

5. **Inductive Hypothesis**:
   - Assume \( f(n - i) \leq 2^i - 1 \) for all \( i \leq m \).

6. **Inductive Step**:
   - For \( f(n - (m + 1)) \):
     \[
     f(n - (m + 1)) \leq \sum_{\ell = n - m}^{n-1} (1 + f(\ell)) \leq 2^0 + 2^1 + \ldots + 2^m = 2^{m+1} - 1
     \]
   - This proves the inductive hypothesis.

7. **Conclusion**:
   - Summing over all contestants:
     \[
     f(1) + f(2) + \ldots + f(n) \leq \sum_{i = 0}^{n-1} (2^i - 1) = 2^n - (n + 1)
     \]
   - Therefore, the process cannot continue indefinitely.

### Part (b): Determine the maximum number of euros the Jury can collect

1. **Initial Claim**:
   - Each contestant \( C_{n-i} \) can make at most \( 2^i - 1 \) moves.

2. **Inductive Construction**:
   - We can turn \( C_n, C_{n-1}, \ldots, C_1 \) into \( C_1, C_2, \ldots, C_n \) legally in \( 2^n - (n+1) \) moves.

3. **Base Case**:
   - For \( n = 0 \), the statement is trivially true.

4. **Inductive Step**:
   - Suppose it is true for some \( m \).
   - For \( m + 1 \):
     \[
     C_{m+1}, C_m, \ldots, C_2, C_1 \to C_{m+1}, C_1, C_2, \ldots, C_m
     \]
     - This takes \( 2^m - (m + 1) \) moves.
     - Then, each \( C_1 \to C_m \) skips over \( C_{m+1} \) to the front of the line, taking \( m \) moves.
     - This turns:
       \[
       C_{m+1}, C_1, C_2, \ldots, C_m \to C_m, C_{m-1}, \ldots, C_1, C_{m+1}
       \]
     - Reverse \( C_m, \ldots, C_1 \) with \( 2^m - (m+1) \) moves to get them into proper order.
     - Total moves:
       \[
       2(2^m - (m+1)) + m = 2^{m+1} - (m+2)
       \]
     - This proves the inductive construction.

### Conclusion
\(\boxed{2^n - (n + 1)}\)