Compliment me

$n^2 + 3n + 2$ is divisible by $2$ for all $n$ (check cases where $n$ is even or odd). Thus it is sufficient to find cases where $3 \mid n^2 + 3n + 2$ .

Factor as $(n+2)(n+1)$ . Then this is not a multiple of $3$ when $3 \nmid n+2$ and $3 \nmid n+1$ , which occurs only when $3 \mid n$ . Thus we want $1991 - \lfloor \frac{1991}{3} \rfloor = 1328$ .

$6 | n^{2} + 3n + 2 ,,,,, 6 | (n + 1) ( n + 2) ,,,,,,,,,,, (n+1)(n+2)\quad \equiv \quad 0\quad (mod\quad 6)\quad then\quad there\quad can\quad be\quad 4\quad cases\quad :\\ case\quad :\quad 1\quad \\ (n+1)\quad \equiv \quad 0\quad (mod\quad 6)\quad and\quad (n+2)\quad \equiv \quad 1\quad (mod\quad 6)\quad \\ further\quad we\quad get\quad n\quad \equiv \quad -1\quad (mod\quad 6)\quad or\quad n\quad \equiv \quad 5\quad (mod\quad 6)\quad means\quad n\quad will\quad be\quad of\quad the\quad form\quad 6k\quad +\quad 5\\ where\quad k\quad is\quad any\quad natural\quad number\quad \\ case\quad :\quad 2\\ (n+2)\quad \equiv \quad 0\quad (mod\quad 6)\quad and\quad then\quad (n+1)\quad \equiv \quad 5\quad (mod\quad 6)\quad \quad n\quad \equiv \quad 4\quad (mod\quad 6)\quad means\quad n\quad will\quad be\quad of\quad the\quad form\quad \\ 6k\quad +\quad 4\quad where\quad k\quad is\quad any\quad natural\quad number\quad \\ case:\quad 3\\ (n+1)\quad \equiv \quad 2\quad (mod\quad 6)\quad and\quad then\quad (n+2)\quad \equiv \quad 3\quad (mod\quad 6)\quad \quad n\quad \equiv \quad 1\quad (mod\quad 6)\quad means\quad n\quad will\quad be\quad of\quad the\quad form\\ 6k\quad +\quad 1\quad where\quad k\quad is\quad any\quad natural\quad number\quad \\ case\quad :\quad 4\\ (n+1)\quad \equiv \quad 3\quad (mod\quad 6)\quad and\quad then\quad (n+2)\quad \equiv \quad 4\quad (mod\quad 6)\quad \quad n\quad \equiv \quad 2\quad (mod\quad 6)\quad means\quad n\quad will\quad be\quad of\quad the\quad form\\ 6k\quad +\quad 2\quad where\quad k\quad is\quad any\quad natural\quad number\\ therefore\quad n\quad will\quad be\quad a\quad number\quad of\quad the\quad form\quad 6k+1\quad or\quad 6k+2\quad or\quad 6k+4\quad or\quad 6k+5\\ for\quad k\quad =\quad 0\quad the\quad numbers\quad obtained\quad are\quad 1,2,4,5\quad now\quad we\quad know\quad that\quad maximum\quad possible\quad value\quad of\quad n\quad =\quad 1991\\ rem(1991\quad \div \quad 6)\quad =\quad 5\quad therefor\quad it\quad is\quad a\quad number\quad of\quad the\quad form\quad 6k\quad +\quad 5\quad and\quad quotient(1991\quad \div \quad 6)\quad =\quad 331\quad so\quad there\quad \\ are\quad 331\quad numbers\quad greater\quad than\quad 6\quad and\quad less\quad than\quad 1992\quad which\quad are\quad of\quad this\quad form\quad and\quad 1\quad number\quad "5"\quad so\quad total\quad 332\quad numbers\quad \\ of\quad this\quad form.\quad \\ similarly\quad we\quad find\quad no.s\quad of\quad the\quad form\quad 6k\quad +\quad 4,\quad 6k\quad +\quad 1\quad ,\quad 6k\quad +\quad 2\quad are\quad all\quad 332\quad each\quad so\quad total \boxed{1328}$