Given: $$a_1=5$$ and $$a_{20}=25$$

Also given, $$a_1, a_2, a_3, \ldots \ldots \ldots$$ are in H.P.

$$\Rightarrow \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \ldots \ldots$$ are in A.P.

Let D be the common difference of above A. P.

$$\begin{array}{ll} \therefore & \frac{1}{a_{20}}=\frac{1}{a_1}+(20-1) d \\ \Rightarrow & \frac{1}{25}=\frac{1}{5}+19 d \\ \Rightarrow & d=\frac{-4}{475} \end{array}$$

$$\begin{aligned} & \text { Now, } \quad \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \\ & \Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}+(n-1) \cdot\left(\frac{-4}{475}\right) \\ & \Rightarrow \quad \frac{1}{a_n}=\frac{95-4 n+4}{475} \\ & \Rightarrow \quad a_n=\frac{475}{99-4 n} \\ \end{aligned}$$

Apply $$\quad a_n<0$$

$$\Rightarrow \quad \frac{475}{99-4 n}<0$$

The least positive integral value of $$n$$ is 25 which satisfy the above condition.