10. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. 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
$$
\left[\frac{1}{S}+\frac{1}{S_{2}}+\cdots+\frac{1}{S_{100}}\right]=(\quad) \text {. }
$$
(A) 17
(B) 18
(C) 19
(D) 20
10. B.

Notice
$$
\begin{array}{l}
S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)=\frac{1}{2}\left[\left(S_{n}-S_{n-1}\right)+\frac{1}{S_{n}-S_{n-1}}\right], \\
S_{n}+S_{n-1}=\frac{1}{S_{n}-S_{n-1}}, \\
S_{n}^{2}=S_{n-1}^{2}+1 .
\end{array}
$$

Since $S_{1}=a_{1}=1$, we have $S_{n}^{2}=n, S_{n}=\sqrt{n}$. Therefore, $\sqrt{n}+\sqrt{n-1}\sqrt{101}-1>9 \Rightarrow S>18 .
\end{array}
$$

Also, since $S_{1}=a_{1}=1$, we have
$$
\begin{array}{l}
\frac{S}{2}-\frac{1}{2 S_{1}}=\frac{1}{2 S_{2}}+\frac{1}{2 S_{3}}+\cdots+\frac{1}{2 S_{100}} \\
<\sqrt{100}-1=9,
\end{array}
$$

which means $S<2\left(9+\frac{1}{2}\right)=19$.
Thus, $[S]=18$.