(10) In the arithmetic sequence $\left\{a_{n}\right\}$, if $\frac{a_{11}}{a_{10}}<-1$, and the sum of the first $n$ terms $S_{n}$ has a maximum value, then when $S_{n}$ takes the smallest positive value, $n=(\quad$ ).
(A) 1
(B) 10
(C) 19
(D) 20
110 $\mathrm{C}$ Hint: Let the common difference of the arithmetic sequence be $d$. Clearly $d<0, a_{11}<0$.
\end{array}
$$

From $a_{11}+a_{10}<0$, we know $2 a_{1}+19 d<0$. Therefore, we have
$$
\begin{aligned}
S_{19}-S_{1} & =19 a_{1}+\frac{19 \times 18}{2} d-a_{1} \\
& =18 a_{1}+9 \times 19 d \\
& =9\left(2 a_{1}+19 d\right)<0 .
\end{aligned}
$$

Thus, $n=19$.