[ Arithmetic progression $]$

The numbers 3, 9, and 15 were written on the board. It was allowed to add two of the written numbers, subtract the third from this sum, and write the result on the board instead of the number that was subtracted. After performing this operation multiple times, three numbers were left on the board, the smallest of which was 2013. What were the other two numbers?
Note that $9-3=6$ and $15-9=6$. We will show that at any moment, one of the numbers on the board will be 6 less than the second and 6 more than the third.

Indeed, suppose this property is satisfied, and the numbers on the board are $x-6, x$, and $x+6$. If we add the two outer numbers and subtract the middle one, the triplet of numbers will not change. If we add the first two numbers and subtract the third, we get the triplet $x-6, x$, and $x-12$, and if we add the last two numbers and subtract the first, we get the triplet $x+12, x$, and $x+6$. In all cases, the specified property is preserved, so it will be satisfied after each step. Therefore, the required numbers are: $2013+6=2019$ and $2019+6=2025$.

## Answer

2019 and 2025.