12. Let $a_{n}$ denote the positive integer closest to $\sqrt{n}$, $n \in \mathbf{N}^{*}$, and set $S=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2000}}$, then $[S]=$
12. 88 Let $a_{n}=k$, then $k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2}, k^{2}-k+\frac{1}{4}<n<k^{2}+k+\frac{1}{4}$. That is, $k^{2}-k+1 \leqslant n \leqslant$ $k^{2}+k$, so the index $n$ for which $a_{n}=k$ is exactly $2 k$. And $45^{2}-45+1=1981<2000<45^{2}+45$, hence $S=2 \times 44+$ $\frac{2000-1981+1}{45}$, which means $[S]=88$.