jmc

Note that $$n^2-10n+29=(n-5{)}^2+4>(n-5{)}^2.$$We also claim that $n^2-10n+29<(n-4{)}^2.$ Expanding, we get $$n^2-10n+29<n^2-8n+16,$$which is equivalent to $2n>13.$ This certainly holds for $n=19941994.$

Hence, $$(n-5{)}^2<n^2-10n+29<(n-4{)}^2,$$so $n-5<\sqrt{n^2-10n+29}<n-4,$ which means that $$\lfloor \sqrt{n^2-10n+29}\rfloor =n-5=\boxed{19941989}.$$