1. On the board, there is a positive integer $n$. In one step, we can erase the number on the board and write either its double, or its double increased by 1. For how many initial numbers $n$ different from 2019 can we achieve that the number 2019 appears on the board after a finite number of steps?

(Josef Tkadlec)
1. In one step, the current natural number $k$ is increased either to the even number $m=2k$, or to the odd number $m=2k+1$. Depending on the parity of the new number $m$, we can reconstruct the previous number $k$: either $k=m/2$, or $k=(m-1)/2$ — depending on whether $m$ is even or odd.

The odd number 2019 will thus appear on the board only after the number $(2019-1)/2=1009$. Since this is again an odd number, after two steps we will reach the target number 2019 only from the number $(1009-1)/2=504$. This is an even number, so after three steps we get 2019 only from the number $504/2=252$, etc. The entire process of determining all suitable numbers from the final 2019 leads to the following result:

$$
2019 \leftarrow 1009 \leftarrow 504 \leftarrow 252 \leftarrow 126 \leftarrow 63 \leftarrow 31 \leftarrow 15 \leftarrow 7 \leftarrow 3 \leftarrow 1 .
$$

(The number 1 is the smallest natural number, so we do not continue further.)

Answer. There are ten such initial values of the number $n$ (they are $1, 3, 7, 15, 31, 63, 126, 252, 504$, and 1009).

Alternative Solution. If we write the numbers on the board in binary, each modification of the current number $n$ consists in not erasing the number at all, but simply appending a zero (changing $n$ to $2n$) or a one (changing $n$ to $2n+1$) to the right of its representation. The number 2019 will thus be obtained after a certain number of steps precisely from such numbers whose binary representation is formed by the first few digits of the binary representation of the number 2019. Since $2^{10}=1024<2019<2048=2^{11}$, the binary representation of the number 2019 has exactly 11 digits, so there are a total of 10 initial numbers $n$ from which we get the number 2019 after one or more steps. (The smallest of these numbers is the number 1, which corresponds to the first digit of the binary representation of the number 2019; the problem does not ask for the other nine numbers.)

Note. The binary representation of the number 2019 itself does not concern us; we usually find it precisely by the sequence of modifications described in the first solution. Thus, the binary representation 11111100011 of the number 2019 is obtained when in the obtained sequence of numbers $1, 3, 7, \ldots, 1009$, 2019 we replace each odd number with a one and each even number with a zero.

For a complete solution, award 6 points, of which 4 points for a general justification that each previous number is determined by the following number, 1 point for finding all numbers down to the unit, and 1 point for formulating the correct conclusion.

The general description in the first paragraph of the solution is not necessary (nor is the final note about the smallest natural number 1); the solver can start the backward process directly by determining the number 1009 from the number 2009 and continue further. However, if they make a numerical error in the step-by-step calculations, do not award more than 4 points. If they make multiple errors, award no more than 3 points. For solutions that found 1009 but did not continue searching for other numbers, award 1 point.

In the case of the second solution, award 1 point for describing the behavior of the operation $\times 2$ and 1 point for describing the behavior of the operation $\times 2+1$ in binary. Another 2 points for explaining which numbers correspond to the sought ones, 1 point for determining the number of digits of the number 2019 in binary (either by estimation or manual conversion), and finally 1 point for the conclusion itself.