Inspired by Huan Le Quang

We start by dividing the polynomial

$\dfrac{n^{2}+11n+2}{n-5}=n+16+\dfrac{82}{n-5}$

Since we want this expression to be an positive integer so $(n-5) | 82$

Then

$n-5=\pm 1,\pm 2, \pm 41, \pm 82$

$n=4,6,3,7,-36,46,-77,87$ but only 6 of these is positive ,so , the answer is $\boxed{6}$

$\frac{ n^2 + 11n + 2 } { n-5}=\frac{ (n+16)(n-5)+82 } { n-5}=n+16+\frac{82}{n-5}$

So we are looking for the number of integers $n$ for which

$n-5\mid82$

which is just the number of positive divisors of $82=2\cdot41$ plus its negative divisors less than $5$ , as $41>5$ , this is just the factors of $2$ and thus out answer is $(1+1)(1+1)+2=\boxed{6}$