Counting Solutions

$n^{2}+3n+2=(n+1)(n+2)$ Observe that if $n$ is not divisible by $3$ , then we get remainder $1$ or $2$ when $n$ is divided by $3$ .
$\Rightarrow$ either $n+1$ or $n+2$ is divisible by $3$ .
$\Rightarrow 6 \mid (n+1)(n+2)$
Since, as $(n+1)(n+2)$ is the product of two consecutive integers, it is also divisible by $2$ .
Thus, for $6$ to be a factor of $n^{2}+3n+2$ , $n$ shouldn't be divisible by $3$ .
In order to find the number of positive integers less than or equal to $1991$ , we use the formula $\left \lfloor{\frac{1991}{3}}\right \rfloor$ $=663$ Thus, the number of positive integers for which the given conditions hold true are $1991-663=\boxed{1328}$