1. All natural numbers from 1 to 2018 are written in a row on the board: 1, 2, 3, ..., 2018. Find any two of them, the removal of which will result in the sum of all numbers between the erased ones being half the sum of all the remaining non-erased numbers? (Problem author - A. Gолованов)
Answer. For example, 673 and 1346. Solution. Note that the sums of numbers equally distant from the ends of the sequence are equal: $1+2018=2+2017=\ldots=1009+1010$. If we erase one such pair of numbers, the number of pairs left will be $1008=336 \cdot 3$. Therefore, if there are 336 pairs between the erased numbers, there will be $336 \cdot 2=672$ pairs left on the outside, and the condition of the problem will be satisfied. This is exactly what happens if we erase the numbers 673 and 1346. Remark. The example given in the answer is not the only one: for instance, the numbers 1289 and 1738 also work.