Let $n$ be a positive integer. On a blackboard, Bobo writes a list of $n$ non-negative integers. He then performs a sequence of moves, each of which is as follows:

-for each $i = 1, . . . , n$, he computes the number $a_i$ of integers currently on the board that are at most $i$,

-he erases all integers on the board,

-he writes on the board the numbers $a_1, a_2,\ldots , a_n$.

For instance, if $n = 5$ and the numbers initially on the board are $0, 7, 2, 6, 2$, after the first move the numbers on the board will be $1, 3, 3, 3, 3$, after the second they will be $1, 1, 5, 5, 5$, and so on.

(a) Show that, whatever $n$ and whatever the initial configuration, the numbers on the board will eventually not change any more.

(b) As a function of $n$, determine the minimum integer $k$ such that, whatever the initial configuration, moves from the $k$-th onwards will not change the numbers written on the board.
### Part (a)
We need to show that, whatever \( n \) and whatever the initial configuration, the numbers on the board will eventually not change anymore.

1. **Initial Observation**: 
   - Let \( a_1, a_2, \ldots, a_n \) be the numbers on the board after the first move.
   - By definition, \( a_i \) is the number of integers on the board that are at most \( i \).

2. **Monotonicity**:
   - Notice that \( a_1 \leq a_2 \leq \ldots \leq a_n \) because the number of integers that are at most \( i \) is always less than or equal to the number of integers that are at most \( i+1 \).

3. **Boundedness**:
   - Each \( a_i \) is a non-negative integer and \( a_i \leq n \) for all \( i \).

4. **Finite States**:
   - Since each \( a_i \) can take on only a finite number of values (from 0 to \( n \)), and there are \( n \) such \( a_i \)'s, the sequence of moves must eventually repeat a previous state.

5. **Convergence**:
   - Once the sequence of numbers repeats, the process will continue to produce the same sequence of numbers, meaning the numbers on the board will not change anymore.

Thus, the numbers on the board will eventually not change anymore.

### Part (b)
We need to determine the minimum integer \( k \) such that, whatever the initial configuration, moves from the \( k \)-th onwards will not change the numbers written on the board.

1. **Base Case**:
   - For \( n = 1 \), the sequence stabilizes in at most 2 moves. If the initial number is 0, it becomes 1 in the next move. If it is 1, it remains 1.

2. **Induction Hypothesis**:
   - Assume that for \( n \leq m \), the sequence stabilizes in at most \( 2m \) moves.

3. **Induction Step**:
   - Consider \( n = m + 1 \).
   - After the first move, all numbers are at most \( n \).
   - After the second move, the last number is \( n \) because \( a_n = n \).
   - The remaining \( n-1 \) numbers are now bounded by \( n-1 \), and by the induction hypothesis, they will stabilize in at most \( 2(n-1) \) moves.
   - Therefore, the entire sequence will stabilize in at most \( 2(n-1) + 2 = 2n \) moves.

Thus, the minimum integer \( k \) such that the sequence stabilizes is \( 2n \).

The final answer is \( \boxed{ 2n } \)