Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$, then Todd must say the next two numbers ($2$ and $3$), then Tucker must say the next three numbers  ($4$, $5$, $6$), then Tadd must say the next four numbers ($7$, $8$, $9$, $10$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$
Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||}   \hline  Round & Tadd & Todd & Tucker \\ [0.5ex]   \hline\hline  1 & 1 & 2-3 & 4-6 \\   \hline  2 & 7-10 & 11-15 & 16-21 \\  \hline  3 & 22-28 & 29-36 & 37-45 \\  \hline  4 & 46-55 & 56-66 & 67-78 \\ [1ex]   \hline \end{tabular}$

Tadd says $1$ number in round 1, $4$ numbers in round 2, $7$ numbers in round 3, and in general $3n - 2$ numbers in round n. At the end of round n, the number of numbers Tadd has said so far is $1 + 4 + 7 + \dots + (3n - 2) = \frac{n(3n-1)}{2}$, by the sum of arithmetic series formula. 
We find that $\dfrac{37(110)}{2}=2035$, so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37. At the end of Tadd's turn in round 37, the children have, in total, completed $36+36+37=109$ turns. In general, at the end of turn $n$, the nth triangular number is said, or $\dfrac{n(n+1)}{2}$. So at the end of turn 109 (or the end of Tadd's turn in round 37), Tadd says the number $\dfrac{109(110)}{2}=5995$. Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was $5995-16=5979$.
Thus, the answer is $\boxed{\textbf{(C) }5979}$.