, 2022

Solution:

From the assumption it follows that the smallest of the numbers $a_{1}, a_{2}, \ldots, a_{2022}$ is at least $1$, the second smallest is at least $2$, and so on up to the $2022$-th, which must be at least $2022$. If we now successively replace the numbers $a_{1}, \ldots, a_{2022}$, in order of size, by $1,2, \ldots, 2022$, then the sum from the problem statement does not increase. We thus see that we may restrict ourselves to the case where $(a_{1}, a_{2}, \ldots, a_{2022})$ are pairwise distinct positive integers. Now we replace $2022$ by an arbitrary positive integer $k$ and claim that the smallest value $s(k)$ that the sum
$$\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{k}}{k}\right\rfloor$$
can take, when $a_{1}, a_{2}, \ldots, a_{k}$ are pairwise distinct positive integers, is exactly $\left\lfloor\log_{2}(k)\right\rfloor$.

- Lower bound: Since $m$ is clearly monotonically increasing, it suffices to show that $m\left(2^{t}\right) \geq t$ holds. The case $t=0$ is clear, for $t>0$ we use complete induction and thus assume $m\left(2^{t-1}\right) \geq t-1$ as known. Let $a_{1}, a_{2}, \ldots, a_{2^{t}}$ be arbitrary pairwise distinct positive integers. Then the maximum $a_{i}$ of these numbers is at least $2^{t}$. If $i>2^{t-1}$, then we have
$$\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{2^{t}}}{2^{t}}\right\rfloor \geq \left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{2^{t-1}}}{2^{t-1}}\right\rfloor+\left\lfloor\frac{a_{i}}{i}\right\rfloor \geq m\left(2^{t-1}\right)+1 = t-1+1 = t$$
If, however, $i \leq 2^{t-1}$, then the set $X=\{1,2, \ldots, 2^{t-1}\} \setminus \{a_{1}, a_{2}, \ldots, a_{2^{t-1}}\}$ is nonempty. Let $b \in X$ be arbitrary. Then $a_{i} \geq 2^{t} = 2^{t-1} + 2^{t-1} \geq b+i$ holds, so that
$$\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{2^{t}}}{2^{t}}\right\rfloor \geq \left\lfloor\frac{a_{1}}{1}\right\rfloor+\cdots+\left\lfloor\frac{b+i}{i}\right\rfloor+\cdots+\left\lfloor\frac{a_{2^{t-1}}}{2^{t-1}}\right\rfloor \geq m\left(2^{t-1}\right)+1 = t-1+1 = t$$

- Upper bound: We set
$$a_{i}=\begin{cases} i-1, & \text{if } i \text{ is not a power of two,} \\ 2^{s+1}-1, & \text{if } i=2^{s}. \end{cases}$$
Then $a_{1}, a_{2}, \ldots, a_{k}$ are pairwise distinct positive integers and we have
$$\left\lfloor\frac{a_{i}}{i}\right\rfloor=\begin{cases} 0, & \text{if } i \text{ is not a power of two,} \\ 1 & \text{otherwise.} \end{cases}$$
hence it follows that $\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{k}}{k}\right\rfloor=\left\lfloor\log_{2}(k)\right\rfloor$.

Applying this to the concrete situation of the problem statement, we obtain that the smallest value that the given sum can take is equal to $\left\lfloor\log_{2}(2022)\right\rfloor=11$.