smc

What we want to know is for how many $n$ is $$\gcd (n^2+7,n+4)>1.$$ We start by setting $$n+4\equiv 0\mathrm{mod}m$$ for some arbitrary $m$. This shows that $m$ evenly divides $n+4$. Next we want to see under which conditions $m$ also divides $n^2+7$. We know from the previous statement that $$n\equiv -4\mathrm{mod}m$$ and thus $$n^2\equiv (-4{)}^2\equiv 16\mathrm{mod}m.$$ Next we simply add $7$ to get $$n^2+7\equiv 23\mathrm{mod}m.$$ However, we also want $$n^2+7\equiv 0\mathrm{mod}m$$ which leads to $$n^2+7\equiv 23\equiv 0\mathrm{mod}m$$ from the previous statement. From that statement, we get that $m$ divides $23$ evenly. Since $23$ is prime and we're looking for a GCD greater than 1, $m$ must be $23$. Going back to our original statement, we can set $$n+4=23x$$ for some positive integer x, and $$n=23x-4.$$ Finally, we must find the largest $x$ such that $$23x-4<1990.$$ This is a simple linear inequality for which the answer is $x=86$, or $\boxed{86}$.