1. In what order should the numbers $1,2, \ldots, 2017$ be rearranged so that for the resulting sequence $a_{1}, a_{2}, \ldots, a_{2017}$, the expression

$$
1 \cdot a_{1}+2 \cdot a_{2}+\ldots+2017 \cdot a_{2017}
$$

has the smallest possible value?

(I. N. Sergeev)

Answer: strictly in reverse order - $2017,2016, \ldots, 1$.
Solution.

A. If the set $a_{1}, a_{2}, \ldots, a_{2017}$ does not coincide with the set $2017,2016, \ldots, 1$, then there will be two indices $i < j$ such that $a_i < a_j$.

B. Since the number of such different sets is finite, the minimum of the given expression is necessarily achieved on some set $a_{1}, a_{2}, \ldots, a_{2017}$, and this set, according to point A, cannot be unordered in descending order.

Note. Without the clarification from point B, the solution is incomplete, as point A only proves that the minimum is not achieved on other sets. What if it is never achieved at all, what if the process of reducing the value of the expression never ends?