(10) For a real number $x$, $[x]$ denotes the greatest integer not exceeding $x$. It is known that the sequence of positive numbers $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)$, where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then $\left[\frac{1}{S_{1}}+\frac{1}{S_{2}}+\cdots+\frac{1}{S_{100}}\right]=(\quad)$.
(A) 17
(B) 18
(C) 19
(D) 20
10
$$
\begin{aligned}
S_{n} & =\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right) \\
& =\frac{1}{2}\left(S_{n}-S_{n-1}+\frac{1}{S_{n}-S_{n-1}}\right),
\end{aligned}
$$

we get
$$
S_{n}+S_{n-1}=\frac{1}{S_{n}-S_{n-1}},
$$

which means $S_{n}^{2}=S_{n-1}^{2}+1$. Since $S_{1}=a_{1}=1$, we have $S_{n}^{2}=n, S_{n}=\sqrt{n}$. Because
$$
\sqrt{n}+\sqrt{n-1}\sqrt{101}-1>9,
$$

solving this gives $S>18$. Also, since $S_{1}=a_{1}=1$, we have
$$
\begin{aligned}
\frac{S}{2}-\frac{1}{2 S_{1}} & =\frac{1}{2 S_{2}}+\frac{1}{2 S_{3}}+\cdots+\frac{1}{2 S_{100}} \\
& <\sqrt{100}-1 \\
& =9
\end{aligned}
$$

which means $S<2\left(9+\frac{1}{2}\right)=19$, thus we get $[S]=18$. Therefore, the answer is B.