Eight zeros are written on a blackboard...

As Gribouillis said the sum is $8n+28$. This is multiple of $4$, but not of $8$. Then, as $9m=8n+28$, using the same argument you used, the possible values are $36+72k$.

Now, it is easy to obtain the solution $99999999$. For example, if we consider the first 4 digits: $0000\rightarrow3321\rightarrow4554\rightarrow7866\rightarrow9999$. (You obtain $99999999$ repeating the same in the other digits).

Then, if you have a solution for $k=0$ (hence, for $12345678$) you will have a solution for $k=1$ (adding up $99999999$). And, by induction, for any $k\geq 0$.

Hence, we have reduced the problem to solve $12345678$. You can try to solve it or see the spoiler below:

\begin{align} 00000000 \ (+ 12330000) \\12330000 \ (+00010233)\\12340233\ (+00002133) \\ 12342366 \ (+00003312) \\ 12345678 \ (+00003312) \end{align}

Pd: This is my first post here, hope everything is clear because my MathJax skills are not the best :)