Answer (C): If $S$ consists of the positive integers less than or equal to $2024$ that are congruent to $1$, $4$, or $8$ modulo $10$, then every pair of elements in $S$ differ by at least $|4 - 1| = |11 - 8| = 3$, and every pair of odd elements of $S$ differ by at least $|11 - 1| = 10$. This set,
$$\{1, 4, 8, 11, 14, 18, \dots, 2011, 2014, 2018, 2021, 2024\},$$
satisfies the given conditions and has $3 \cdot (\frac{2020}{10}) + 2 = 608$ elements. To see that no larger set satisfies the given conditions, note that if a set satisfies the first condition and some block of $10$ consecutive integers contains $4$ elements of the set, then those $4$ elements would need to be the $1$st, $4$th, $7$th, and $10$th elements in that block, and the two odd numbers among them would differ by $6$, in violation of the second condition. Therefore there are at most $3 \cdot 202 = 606$ elements of $S$ among the first $2020$ positive integers, and at most $2$ elements of $S$ can be among $\{2021, 2022, 2023, 2024\}$.