$k = 1011$

Let $b_i = \frac{a_1 + a_2 + \ldots + a_i}{1 + 2 + \ldots + i}$. Now, note that $k = 1011$ is achievable by letting $a_i = 2i~~\forall i \leq 1011$ and completing the permutation in any way. This will have $b_i = 2$ for $i \leq 1011$. It remains to show the upper bound. I claim that at most 1 $i \geq 1011$ can have $b_i$ be an integer larger than 1. First note that $b_i \leq \frac{n + (n-1) + \ldots + (n-i+1)}{1 + 2 + \ldots + i} = \frac{n + n - i + 1}{1 + i} \leq \frac{3034}{1012} < 3$ for $i \geq 1011$, so the only possible value of $b_i$ is 2. Now note that if $b_i \leq 2$ then $b_{i+1} < 2$. We can see this because $\frac{a_{i+1}}{i+1} \leq \frac{2022}{i+1} \leq \frac{2022}{1012} < 2$ and as $b_{i+1}$ is the mediant of $b_i$ and $\frac{a_{i+1}}{i+1}$ this implies $b_{i+1} < 2$. This means that if $b_i = 2$ for any $i \geq 1011$ then all further $b_j, j > i$ will be less than $2$ and cannot be integers. Thus the maximum number of $b_i$ that are integers larger than $1$, is the first $1010$ and at most $1$ from the remaining terms. Thus $k \leq 1011$ as desired.