Notice that $84 = 4 \times 3 \times 7$. Let $M$ be the resulting number after Lionel erases some digits. We want $M$ to be even, so we must erase the rightmost $3$. We also want $M$ to be divisible by $3$. For this to happen, the sum of digits of $M$ must be divisible by $3$. Initially, the sum of digits equals $100 \cdot (2 + 0 + 2 + 3) = 700 \equiv 1 \pmod{3}$. Since erasing digits $0$ or $3$ does not alter the congruence class modulo $3$, we must erase some digits $2$. If we erase $x$ of these digits, then $M$ will be equivalent to $700 - 2x$ modulo $3$. This number is divisible by $3$ if and only if $x \equiv 2 \pmod{3}$, hence $x \ge 2$, i.e., we must remove at least two digits $2$.

So far, we have proved that Lionel needs to remove at least two digits $2$ and one digit $3$. Suppose it is possible to erase only those three digits, and let us now think about how to obtain the largest possible multiple of $84$ in this situation. To obtain a multiple of $4$, the last digit $2$ should be removed (otherwise, the last two digits of $M$ would be $02$). So we have the following number, in which we must erase one more digit $2$:
$$[2023][2023]\dots[2023]20.$$
We will not erase the last digit $2$, as $M$ would end up not being divisible by $4$. We classify the rest of the $2$'s in two groups: those that are the first digit of a $2023$ block and those that are the third digit of a $2023$ block. If we remove one of the latter, then $M$ will have the form
$$[2023]\dots[2023][203][2023]\dots[2023]20.$$
Since $2023$ and $203$ are divisible by $7$ and $20$ is not, this number is not divisible by $7$. Hence we must remove the leading digit of some $2023$ block. In order to obtain the highest possible number, it is always preferable to choose a block from the right, since $0232023 < 2023023$. We can see by direct inspection that erasing the leading digit of either of the two last blocks does not work (because the resulting number is not divisible by $7$), whereas choosing the third rightmost block does work. So Lionel's final number is:
$$M = [2023][2023]\dots[2023][023][2023][2023]20.$$