Rioplatense Mathematical Olympiad

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 $100\cdot (2+0+2+3)=700\equiv 1(\mathrm{mod}3)$. 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 $x\equiv 2(\mathrm{mod}3)$, 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: $$M=[2023][2023]\dots [2023][023][2023][2023]20.$$