Solution:

We call a configuration stable if the numbers written on the board are exactly the numbers from 1 to $n$, each appearing once, and we observe that such a configuration no longer changes when Bobo applies the procedure described in the statement.

We show that $2n$ moves always suffice to reach the stable configuration, whatever the initial configuration, and that $2n$ moves can indeed be needed for suitable choices of the initial configuration (and hence $k=2n+1$).

$2n$ moves suffice to reach the stable configuration. First of all, observe that the numbers computed during any move always satisfy the relations
$$0 \leq a_{1} \leq a_{2} \leq \ldots \leq a_{n} \leq n$$
Indeed $a_{i} \leq a_{i+1}$ for every admissible index $i$ (this is because if $x \leq i$, then $x \leq i+1$), and there are $n$ numbers on the board, so $a_{n} \leq n$. For simplicity, we assume that after each move Bobo writes the numbers $a_{1}, \ldots, a_{n}$ on the board in increasing order.

By what we have just remarked, from the second move onward, the last number written on the board is always $n$ (that is, one computes $a_{n}=n$). In carrying out the third move, since the numbers written on the board are integers at most $n$ and the last one is $n$, one will compute $a_{n-1} \leq n-1$. But then, since the numbers written after the third move satisfy the relations
$$0 \leq a_{1} \leq a_{2} \leq \ldots \leq a_{n-1} \leq n-1 < a_{n}$$
in carrying out the fourth move one finds $a_{n-1}=n-1$, and this holds from then on. In the computation of the fifth move we will have $a_{n-2} \leq n-2$ (since the list at this point contains $n$ and $n-1$), and in the sixth move $a_{n-2}=n-2$.

More generally, after move $2h$ the last numbers written on the board will be the numbers from $n-h+1$ to $n$, so in move $2h+1$ we will compute $a_{n-h} \leq n-h$, and we will write exactly $n-h$ numbers less than or equal to $n-h$. Hence in move $2h+2$ the last numbers written will be $n-h, n-h+1, \ldots, n$.

After move $2n$, the numbers written will be $1,2,3, \ldots, n$, and they will remain so from then on.

$2n$ moves can be needed to reach the stable configuration. Suppose that the initial numbers are all strictly greater than $n$. We will show that in this case $2n$ moves are necessary to reach the stable configuration. When $n=1$ this is an easy check (after one move the board has the number 0, after two moves the number 1).

In general, after the first move, the numbers on the board will all be 0, and hence after the second move they will all be equal to $n$. As in the first part of the proof, in all subsequent moves one will compute $a_{n}=n$, while the computation of $a_{i}$ for $i<n$ is not influenced by the presence of $a_{n}=n$ (since $n$ is not $\leq i$ for any $i<n$). In particular, the game proceeds acting only on the $n-1$ smallest among the $n$ numbers present, and in the situation after the second move these $n-1$ numbers are all equal to $n$.

Let us then consider only the numbers $a_{1}, \ldots, a_{n-1}$, ignoring $a_{n}$ (which from now on will always remain equal to $n$). The game continues as if it involved only $n-1$ numbers, which are initially all greater than $n-1$ (since they are equal to $n$). We then repeat the same analysis just carried out: after one move all these numbers will be replaced by 0, and after two moves they will all be equal to $n-1$. From this point on, for the same reason as before, we will always have $a_{n-1}=n-1$, and hence the game will continue as if it involved only $a_{1}, \ldots, a_{n-2}$ (which will first all become zeros, and then all $n-2$). In particular, after 2 moves the number $a_{n}$ becomes equal to $n$ for the first time (and then never changes again), after 4 moves the number $a_{n-1}$ becomes equal to $n-1$ for the first time (and then never changes again), and so on: after $2h$ moves, the number $a_{n+1-h}$ becomes equal to $n+1-h$ for the first time, and then never changes again. The number $a_{1}=a_{n+1-n}$ then becomes equal to 1 for the first time after $2n$ moves: hence $2n$ moves are indeed necessary to reach the stable configuration.