Solution:

The answer is (D). Since the sum of 4 consecutive terms is constant, for every $n \geq 1$ we have $a_{n}+a_{n+1}+a_{n+2}+a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}$, from which $a_{n}=a_{n+4}$, that is, the sequence is periodic and repeats after 4 terms.

Since $a_{1}<a_{2}<a_{3}$ holds, the differences $a_{2}-a_{1}$ and $a_{3}-a_{2}$ are positive and, by hypothesis, equal to a constant $d$. Since $a_{2}=6$, we can write $a_{1}=6-d$, $a_{3}=6+d$ and we obtain $a_{5}=a_{1}=6-d$ by periodicity; $a_{4}-a_{3}= \pm d$, so $a_{4}=6+2 d$ or $a_{4}=6$. In the first case we would have $a_{4}-a_{5}=3 d$, absurd, so necessarily $a_{4}=6$.

The sequence is therefore $6-d, 6,6+d, 6,6-d, 6 \ldots 6-d, 6,6+d$ (since the terms are 2019, which gives remainder 3 when divided by 4). In each block of 4 and in the last block of 3 the elements $6-d$ and $6+d$ have sum $12=6 \cdot 2$ and hence the total sum is $2019 \cdot 6=12114$ independently of $d$.