6. The terms of the sequence $\left\{a_{n}\right\}$ are distinct positive numbers, and their reciprocals form an arithmetic sequence. Then
$$
\frac{a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2014} a_{2015}}{a_{1} a_{2015}}=
$$
$\qquad$
6.2014.
Let the common difference of the reciprocal sequence be $d(d \neq 0)$.
From the condition, we have
$$
\begin{array}{l}
\frac{1}{a_{2}}-\frac{1}{a_{1}}=\frac{1}{a_{3}}-\frac{1}{a_{2}}=\cdots=\frac{1}{a_{2015}}-\frac{1}{a_{2014}}=d \\
\Rightarrow \frac{a_{1}-a_{2}}{a_{1} a_{2}}=\frac{a_{2}-a_{3}}{a_{2} a_{3}}=\cdots=\frac{a_{2014}-a_{2015}}{a_{2014} a_{2015}}=d .
\end{array}
$$

According to the property of the ratio of sums, we get
$$
\frac{a_{1}-a_{2015}}{a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2014} a_{2015}}=d \text {. }
$$

From equation (1), we know
$$
\begin{array}{l}
\frac{1}{a_{2015}}-\frac{1}{a_{1}}=2014 d \\
\Rightarrow \frac{a_{1}-a_{2015}}{a_{1} a_{2015}}=2014 d .
\end{array}
$$

From equations (2) and (3), we get
$$
\frac{a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2014} a_{2015}}{a_{1} a_{2015}}=2014
$$