Integers!

Note that ${ n }^{ 2 }-1$ is divisible by $n-1$ . Thus:

$\frac { { n }^{ 2 }-9 }{ n-1 } =\frac { { n }^{ 2 }-1 }{ n-1 } -\frac { 8 }{ n-1 }$ $=n+1-\frac { 8 }{ n-1 }$

So, if $n$ is an integer, then $\frac { { n }^{ 2 }-9 }{ n-1 }$ is an integer if and only if $n-1$ divides exactly into 8.

The possible values for $n$ are -7, -3, -1, 0, 2, 3, 5, 9. The sum of thes values is $\boxed { 8 }$ .