Once You Get Hold Of It, It's Easy

Firstly let ${x}$ = ${n}$ + 7 to make the equation easier to deal with.... $\frac{(x-6)^2}{x}\ = \frac{x^2 -12x +36}{x}\ = x - 12 + \frac{36}{x}$ , so all we need to do is find the positive divisors of 36 which are $x= 1,2,3,4,6,9,12,18,36\Rightarrow\ n= 2,5,11,29\therefore\displaystyle \sum_{}^\ n = 47$

We have to find all $n$ such that

$n + 7 \mid (n + 1)^2$

$n + 7 \mid n^2 + 2n + 1$

$n + 7 \mid n(n + 7) - 5n + 1$

$n + 7 \mid -5n + 1$

$n + 7 \mid -5(n + 7) + 36$

$n + 7 \mid 36$

We have to now find all factors of 36 greater than 7. These are 9, 12, 18 and 36. The values of $n$ will therefore be 2, 5, 11 and 29. $2 + 5 + 11 + 29 = 47$

$n^2+2n+1=(n-5)(n+7)+36$

$\frac{n^2+2n+1}{7}=(n-5)+\frac{36}{n+7}$

We have to find all possitive $n$ sucha that $n+7|36$

$n+7$ any divisor of $36$ and $n>7$

Divisors of $36$ greater tha $7$ are $9,12,18,36 \therefore n=2,5,11,29$

Sum is $2+5+11+29=\boxed{47}$

Inevitably, similar to the solutions by Messrs Clement and Kesa; using the Python sympy library.

Synthetic division, then n-7 must divide the remainder. Sum over the positive possibilities.