*Answer:* (a) No; (b) No.

a. The difference and the sum of two integers have equal parity, likewise are an integer and its absolute value of equal parity. Thus Mari's every step keeps the parity of the sum of all numbers on the blackboard unchanged. Among $1, 2, \ldots, 2021$, there are $1011$ odd numbers. As $1011$ is odd itself, the sum of all numbers $1, 2, \ldots, 2021$ is odd. This implies that the sum of all integers from $1$ to $2021$ that contain $4$ and the sum of integers from $1$ to $2021$ that do not contain $4$ have different parities. Hence the set of all natural numbers from $1$ to $2021$ that contain $4$ can never appear on the blackboard.

b. Let there be $a$ positive integers from $1$ to $10000$ that do not contain $4$. Then there are $10000 - a$ positive integers from $1$ to $10000$ that contain $4$. As each Mari's step decreases the number of numbers on the blackboard by $1$, she should make $a - (10000 - a)$ or, equivalently, $2a - 10000$ steps to reach the desired state. As only one new number can appear at each step, introducing $10000 - a$ new numbers assumes $2a - 10000 \ge 10000 - a$ which is equivalent to $a \ge \frac{20000}{3}$. But there are only $9^4$ numbers among the first $10000$ positive integers that do not contain $4$; since $9^4 = 6561 < \frac{20000}{3}$, achieving the desired state is impossible.