10.2. In a row, 21 numbers are written sequentially: from 1999 to 2019 inclusive. Enthusiastic numerologists Vova and Dima performed the following ritual: first, Vova erased several consecutive numbers, then Dima erased several consecutive numbers, and finally, Vova erased several consecutive numbers (at each step, they erased consecutive natural numbers, without skipping over the gaps that formed). In the end, the sum of the numbers erased by Vova turned out to be exactly four times the sum of the numbers erased by Dima, and only one number remained in the sequence. What number was left unerased?
Solution. Let $n$ be the sum of the numbers erased by Dima, then $4n$ is the sum of the numbers erased by Vova, $5n$ is the sum of the numbers erased by both, and $42189 - 5n$ is the number that remains unerased (42189 is the sum of all numbers from 1999 to 2019). According to the problem, $1999 \leq 42189 - 5n \leq 2019$, from which we get $8034 \leq n \leq 8038$. It is clear that Dima erased 4 numbers, which are consecutive by the problem's condition; $8034 = 2007 + 2008 + 2009 + 2010$, $8038 = 2008 + 2009 + 2010 + 2011$, and the intermediate values of $n$ are not sums of four consecutive numbers. When $n = 8034$, we get $5n = 40170$, $42189 - 5n = 2019$, and when $n = 8038$, we get $5n = 40190$, $42189 - 5n = 1999$ - both cases are consistent, and in either case, the unerased number is one of the extreme numbers.

Answer: 1999 or 2019.

Solutions to the mathematics olympiad problems for the 2019-2020 academic year, 10th grade