jmc

Note that $f(k^2)=k^4+k^2+1.$ By a little give and take, $$\begin{array}{rl}f(k^2)& =(k^4+2k^2+1)-k^2\\ & =(k^2+1{)}^2-k^2\\ & =(k^2+k+1)(k^2-k+1)\\ & =f(k)(k^2-k+1).\end{array}$$Furthermore, $$f(k-1)=(k-1{)}^2+(k-1)+1=k^2-2k+1+k-1=k^2-k+1,$$so $$f(k^2)=f(k)f(k-1).$$Thus, the given inequality becomes $$1000f(1)f(0)\cdot f(2)f(1)\cdot f(3)f(2)\cdots f(n-1)f(n-2)\cdot f(n)f(n-1)\ge f(1{)}^{2}f(2{)}^2\cdots f(n{)}^2,$$which simplifies to $$1000\ge f(n).$$The function $f(n)$ is increasing, and $f(31)=993$ and $f(32)=1057,$ so the largest such $n$ is $\boxed{31}.$