Steve wrote the digits $1$, $2$, $3$, $4$, and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$rd, $6$th, $9$th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$th, $8$th, $12$th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$
Note that cycles exist initially and after each round of erasing.
Let the parentheses denote cycles. It follows that:


Initially, the list has cycles of length $5:$ \[(12345)=12345123451234512345\cdots.\]
To find one cycle after the first round of erasing, we need one cycle of length $\operatorname{lcm}(3,5)=15$ before erasing. So, we first group $\frac{15}{5}=3$ copies of the current cycle into one, then erase:
\begin{align*} (12345)&\longrightarrow(123451234512345) \\ &\longrightarrow(12\cancel{3}45\cancel{1}23\cancel{4}51\cancel{2}34\cancel{5}) \\ &\longrightarrow(1245235134). \end{align*}
As a quick confirmation, one cycle should have length $15\cdot\left(1-\frac{1}{3}\right)=10$ at this point.
To find one cycle after the second round of erasing, we need one cycle of length $\operatorname{lcm}(4,10)=20$ before erasing. So, we first group $\frac{20}{10}=2$ copies of the current cycle into one, then erase:
\begin{align*} (1245235134)&\longrightarrow(12452351341245235134) \\ &\longrightarrow(124\cancel{5}235\cancel{1}341\cancel{2}452\cancel{3}513\cancel{4}) \\ &\longrightarrow(124235341452513). \end{align*}
As a quick confirmation, one cycle should have length $20\cdot\left(1-\frac{1}{4}\right)=15$ at this point.

To find one cycle after the third round of erasing, we need one cycle of length $\operatorname{lcm}(5,15)=15$ before erasing. We already have it here, so we erase:
\begin{align*} (124235341452513)&\longrightarrow(1242\cancel{3}5341\cancel{4}5251\cancel{3}) \\ &\longrightarrow(124253415251). \end{align*}
As a quick confirmation, one cycle should have length $15\cdot\left(1-\frac{1}{5}\right)=12$ at this point.

Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: \[(12\underline{425}3415251).\] Therefore, the answer is $4+2+5=\boxed{\textbf{(D)}\ 11}.$
~MRENTHUSIASM (inspired by TheBeautyofMath)