By hypothesis, $a$ satisfies
\begin{equation}
N \frac{1}{a} = \sum_{n = 1}^N \frac{1}{a} = \sum_{n = 1}^N \frac{1}{x_n}
\end{equation}
Diving each side by the positive integer $N \geq 1$, we obtain
\begin{equation}
\frac{1}{a}
= \frac{1}{N} \sum_{n = 1}^N \frac{1}{x_n}
= \frac{\sum_{n = 1}^N \frac{1}{x_n}}{N}
\end{equation}
Since neither side is zero, we may invert both sides to conclude with
\begin{equation}
a
= \frac{1}{1/a}
= \frac{1}{\frac{\sum_{n = 1}^N \frac{1}{x_n}}{N}}
= \frac{N}{\sum_{n = 1}^N \frac{1}{x_n}}
\end{equation}
$\square$
