Let $a_{n}$ denote the integer closest to $\sqrt{n}$. What is the value of

$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{1994}}
$$
Consider the sum up to $\frac{1}{a_{m}}$ in general. By definition, $a_{n}=k$ if $k-\frac{1}{2}0$, so

$$
x_{0}=\frac{-1+\sqrt{4 m-3}}{2}
$$

and $p \leq x_{0}<p+1$ implies

$$
p=\left[x_{0}\right]=\left[\frac{-1+\sqrt{4 m-3}}{2}\right]
$$

The sum of the reciprocals can then be written as follows:

$$
\begin{gathered}
\underbrace{2+2+\cdots+2}_{p \text { terms }}+\underbrace{\frac{1}{p+1}+\cdots+\frac{1}{p+1}}_{m-2(1+2+\cdots+p)=m-p(p+1)}=2 p+\frac{m}{p+1}-\frac{p(p+1)}{p+1}= \\
=p+\frac{m}{p+1}=\left[\frac{-1+\sqrt{4 m-3}}{2}\right]+\frac{m}{\left[\frac{-1+\sqrt{4 m-3}}{2}\right]+1}
\end{gathered}
$$

For the case $m=1994$, the sum is $\frac{3974}{45}$.

Based on the work of Gyula Pap (Debrecen, Fazekas M. Gymnasium, 1st year)