Every positive integer is marked with a number from the set $\{ 0,1,2\}$, according to the following rule:
$$\text{if a positive integer }k\text{ is marked with }j,\text{ then the integer }k+j\text{ is marked with }0.$$
Let $S$ denote the sum of marks of the first $2019$ positive integers. Determine the maximum possible value of $S$.

[i]Proposed by Ivan Novak[/i]
1. **Define the variables and constraints:**
   Let \( x_1 \) be the number of integers marked with 2, \( x_2 \) be the number of integers marked with 1, and \( x_3 \) be the number of integers marked with 0. We know that:
   \[
   x_1 + x_2 + x_3 = 2019
   \]
   We aim to maximize the sum \( S = 2x_1 + x_2 \).

2. **Analyze the marking rule:**
   According to the problem, if an integer \( k \) is marked with \( j \), then \( k + j \) is marked with 0. This implies that for every integer marked with 2, the integer two places ahead must be marked with 0, and for every integer marked with 1, the integer one place ahead must be marked with 0.

3. **Establish the relationship between \( x_1 \) and \( x_3 \):**
   For every integer \( j \) marked with 2, \( j + 2 \) must be marked with 0. Therefore, the number of integers marked with 2 cannot exceed the number of integers marked with 0 by more than 2. This gives us:
   \[
   x_1 \leq x_3 + 2
   \]

4. **Substitute \( x_3 \) in terms of \( x_1 \) and \( x_2 \):**
   From the equation \( x_1 + x_2 + x_3 = 2019 \), we can express \( x_3 \) as:
   \[
   x_3 = 2019 - x_1 - x_2
   \]
   Substituting this into the inequality \( x_1 \leq x_3 + 2 \), we get:
   \[
   x_1 \leq 2019 - x_1 - x_2 + 2
   \]
   Simplifying, we find:
   \[
   2x_1 + x_2 \leq 2021
   \]

5. **Maximize \( S = 2x_1 + x_2 \):**
   The maximum value of \( 2x_1 + x_2 \) is 2021, which can be achieved if \( x_1 = 1009 \) and \( x_2 = 3 \). This satisfies the constraint \( x_1 + x_2 + x_3 = 2019 \) with \( x_3 = 1007 \).

6. **Verify the marking pattern:**
   To achieve this, we can mark the integers in the following pattern: 2200|210|210|...|210|22, where the number of 210 sequences is 671. This ensures that the marking rule is followed and the sum of the marks is maximized.

\(\blacksquare\)

The final answer is \( \boxed{2021} \).