252. Let $a_{n}$ denote the integer closest to $\sqrt{n}$. Find the sum

$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{1980}}
$$
252. Answer: 88.

Each number $k=1,2,3, \ldots$ appears in the sequence $(a_{n})$ $2 k$ times, since the condition $a_{n}=k$ is equivalent to

$$
k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2}, \quad \text { or } \quad k^{2}-k<n \leqslant k^{2}+k
$$

Therefore, in the sum

$$
\begin{aligned}
&\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{3}}+\frac{1}{a_{4}}+\frac{1}{a_{5}}\right.\left.+\frac{1}{a_{6}}\right)+\ldots \\
& \ldots+\left(\frac{1}{a_{44 \cdot 43+1}}+\ldots+\frac{1}{a_{44 \cdot 45}}\right)
\end{aligned}
$$

each of the 44 round brackets equals $2 k \cdot 1 / k=2$.