6. For a positive integer $n$, let $a_{n}$ be the integer closest to $\sqrt{n}$. Then $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{200}}=(\quad)$.
(A) $\frac{191}{7}$
(B) $\frac{192}{7}$
(C) $\frac{193}{7}$
(D) $\frac{194}{7}$
6. A.

Notice that, for any natural number $k$,
$$
\left(k+\frac{1}{2}\right)^{2}=k^{2}+k+\frac{1}{4}
$$

is not an integer.
Therefore, for any positive integer $n$, $\sqrt{n}-\frac{1}{2}$ is definitely not an integer.
Let $m$ be the integer closest to $\sqrt{n}$.
Then $|m-\sqrt{n}|<\frac{1}{2}(m \geqslant 1)$.
It is easy to see that, when $m \geqslant 1$,
$$
\begin{array}{l}
|m-\sqrt{n}|<\frac{1}{2} \Leftrightarrow\left(m-\frac{1}{2}\right)^{2}<n<\left(m+\frac{1}{2}\right)^{2} \\
\Leftrightarrow m^{2}-m+\frac{1}{4}<n<m^{2}+m+\frac{1}{4} .
\end{array}
$$

Then for a fixed positive integer $m$, when the positive integer $n$ satisfies
$$
m^{2}-m+1 \leqslant n \leqslant m^{2}+m
$$

$m$ is the integer closest to $\sqrt{n}$, i.e., $a_{n}=m$.
Thus, the number of positive integers $n$ for which $a_{n}=m$ is $2m$. Notice that,
$$
13^{2}+13=182<200<14^{2}+14=210 \text {. }
$$

Therefore, in $a_{1}, a_{2}, \cdots, a_{200}$, there are 2 ones, 4 twos,
$$
\begin{array}{l} 
6 \text { threes, } 8 \text { fours, } \cdots \cdots 26 \text { thirteens, } 18 \text { fourteens. } \\
\text { Hence } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{200}} \\
\quad= 2 \times \frac{1}{1}+4 \times \frac{1}{2}+\cdots+26 \times \frac{1}{13}+18 \times \frac{1}{14} \\
=\frac{191}{7} .
\end{array}
$$