jmc

First let's solve $|n-3|<9$. The absolute value of a quantity is less than 9 if and only if the quantity is between $-9$ and 9, so solve \begin{array}{r@{\;\;<\;\;}c@{\;\;<\;\;}lc} -9 & n-3 & 9 &\quad \implies \\ -9+3 & n & 9+3 &\quad \implies \\ -6 & n & 12. \end{array} Now consider $|n|<|n-3|$. The distance from $n$ to 0 is $|n|$, and the distance from $n$ to 3 is $|n-3|$. Therefore, this inequality is satisfied by the numbers that are closer to 0 than to 3. These are the numbers less than $1.5$. So the integer solutions of $|n|<|n-3|<9$ are $-5$, $-4$, $-3$, $-2$, $-1$, 0, and 1, and their sum is $-5-4-3-2=\boxed{-14}$.