When does $n-4$ divide $3n-17$?

Reduction. We have
$$3n-17=3(n-4)-5.$$
So $n-4\mid 3n-17$ if and only if $n-4\mid 5$.

Case analysis. The divisors of $5$ are $\pm1,\pm5$. Hence
$$n-4\in\{-5,-1,1,5\},$$
giving
$$n\in\{-1,3,5,9\}.$$

Verification. For $n=-1$: $n-4=-5$ and $3n-17=-20$, and $-5\mid-20$. For $n=3$: $-1\mid-8$. For $n=5$: $1\mid-2$. For $n=9$: $5\mid10$. All work.

Remark. This is the standard technique for divisibility of linear expressions: reduce the dividend modulo the divisor to obtain a constant remainder.