jmc

To say that $k=(a_3{a}_2{a}_1{a}_0{)}_{-3+i}$ is to say that $$k=a_3(-3+i{)}^3+a_2(-3+i{)}^2+a_1(-3+i)+a_0.$$Expanding the right-hand side, we have $$k=(-18a_3+8a_2-3a_1+a_0)+(26a_3-6a_2+a_1)i.$$Since $k$ is a real number, the imaginary part of the right-hand side must be zero; that is, $$26a_3-6a_2+a_1=0$$or $$26a_3=6a_2-a_1.$$Remember that $0\le a_1,a_2,a_3\le 9$, so $6a_2-a_1\le 6\cdot 9-0=54$. Thus, $26a_3\le 54$, so $a_3\le 2$. We take cases, remembering that $a_3\ne 0$:

If $a_3=1$, then we have $6a_2-a_1=26$. The only solution to this equation is $(a_1,a_2)=(4,5)$, so we have $$k=-18a_3+8a_2-3a_1+a_0=-18\cdot 1+8\cdot 5-3\cdot 4+a_0=10+a_0.$$Since $a_0\in \{0,1,2,\dots ,9\}$, the possible values of $k$ are $10,11,12,\dots ,19$, and these have a sum $$10+11+12+\cdots +19=\frac{29\cdot 10}{2}=145.$$ If $a_3=2$, then we have $6a_2-a_1=52$. The only solution to this equation is $(a_1,a_2)=(2,9)$, so we have $$k=-18a_3+8a_2-3a_1+a_0=-18\cdot 2+8\cdot 9-3\cdot 2+a_0=30+a_0.$$Therefore, the possible values of $k$ are $30,31,32,\dots ,39$, which sum to $$30+31+32+\cdots +39=\frac{69\cdot 10}{2}=345.$$

Adding up both cases, we get the answer, $145+345=\boxed{490}$.