jmc

First we write down the equation $a_{n+3}=a_{n+2}+a_{n+1}+a_n$ for $n=1,2,3,\dots ,27$: $$\begin{array}{rl}{a}_4& =a_3+a_2+a_1,\\ a_5& =a_4+a_3+a_2,\\ a_6& =a_5+a_4+a_3,\\ ⋮& \\ a_{30}& =a_{29}+a_{28}+a_{27}.\end{array}$$Let $S=a_1+a_2+\dots +a_{28}$ (the desired quantity). Summing all these equations, we see that the left-hand side and right-hand side are equivalent to $$S+a_{29}+a_{30}-a_1-a_2-a_3=(S+a_{29}-a_1-a_2)+(S-a_1)+(S-a_{28}).$$Simplifying and solving for $S$, we obtain $$S=\frac{a_{28}+a_{30}}{2}=\frac{6090307+20603361}{2}=\frac{\dots 3668}{2}=\dots 834.$$Therefore, the remainder when $S$ is divided by $1000$ is $\boxed{834}$.