Question 129
Determine the smallest value that the expression
$$
\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{2022}}{2022}\right\rfloor
$$
can take, where $a_{1}, a_{2}, \ldots, a_{2022} \geq 1$ are real numbers such that $\left|a_{i}-a_{j}\right| \geq 1$ for all $1 \leq i\lt j \leq 2022$.
Note: For a real number $x$, $\lfloor x\rfloor$, called the floor of $x$, denotes the largest integer not greater than $x$.
$$
\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{2022}}{2022}\right\rfloor
$$
can take, where $a_{1}, a_{2}, \ldots, a_{2022} \geq 1$ are real numbers such that $\left|a_{i}-a_{j}\right| \geq 1$ for all $1 \leq i\lt j \leq 2022$.
Note: For a real number $x$, $\lfloor x\rfloor$, called the floor of $x$, denotes the largest integer not greater than $x$.