23. Vise has to take care of three children, A, B, and C. Since Vise knows that the children like large numbers, he designed a counting game for them: First, A says 1, then B must say the next two numbers $2, 3$, C must say the next three numbers $4, 5, 6$, A must say the next four numbers 7, $8, 9, 10, \cdots \cdots$ This process continues cyclically among the three children, with each child saying one more number than the previous child, until they reach 10000. What is the 2019th number that A says?
(A) 5743
(B) 5885
(C) 5979
(D) 6001
(E) 6011
23. C.

First, count the number of numbers each of 甲, 乙, and 丙 says each time:
甲: $1,4,7, \cdots$;
乙: $2,5,8, \cdots$;
丙: $3,6,9, \cdots$.
Thus, the total number of numbers 甲 says in the first $n$ times is
$$
1+4+\cdots+3 n-2=\frac{n(3 n-1)}{2} \text {. }
$$

According to the problem,
$$
\begin{array}{l}
\frac{(k-1)(3(k-1)-1)}{2} \leqslant 2019 \leqslant \frac{k(3 k-1)}{2} \\
\Rightarrow k=37 . \\
\text { Then } \frac{36(3 \times 36-1)}{2}=1926 \text { is the last number 甲 says in the } 36
\end{array}
$$

round, and 2019-1926=93. Therefore, 2019 is the 93rd number 甲 says in the 37th round. In the first 36 rounds, the three people together say
$$
1+2+\cdots+36 \times 3=\frac{108(1+108)}{2}=5886 \text {. }
$$

Thus, the 2019th number 甲 says is
$$
5886+93=5979 \text {. }
$$