NMTC Tricky problems 2

Note that if $f(n)=k,\ 0\le n-k^2<(k+1)^2-n\implies 2k^2+2k+1>2n\implies k^2+k=k(k+1)\ge n$ . Also, since $f(n)\ne k-1\implies 2n>2(k-1)^2+2(k-1)+1\implies n\ge (k-1)^2+(k-1)+1=k(k-1)+1$ . Thus, for a given integer $k,\ k\ge 1$ there are $2k$ integers $n$ such that $f(n)=k$ .Noting that $f(2008)=45$ , and that $45(45-1)+1=1981$ , we get $\sum_{k=1}^{2008}\frac{1}{f(n)}=\sum_{k=1}^{44}\frac{2k}{k}+\frac{2008-1980}{44}=\boxed{88.6222}$