6. There exists a pair of positive integers $(a, b)$ satisfying
$$
\frac{k}{2015}(a+b)=[a, b]
$$

The number of positive integers $k$ is $\qquad$ ( $[a, b]$ denotes the least common multiple of the positive integers $a$ and $b$).
6. 1007.

Let $d=(a, b), a=a_{1} d, b=b_{1} d$.
Then $\left(a_{1}, b_{1}\right)=1$, and $k=\frac{2015 a_{1} b_{1}}{a_{1} b_{1}}$.
Since $\left(a_{1}+b_{1}, a_{1} b_{1}\right)=1$, $a_{1}+b_{1}$ is a divisor of $2015=5 \times 13 \times 31$, and $a_{1}+b_{1}>1$.

Thus, the possible values of $a_{1}+b_{1}$ are $5,13,31$, $5 \times 13,5 \times 31,13 \times 31,5 \times 13 \times 31$.
It is easy to see that when $a_{1}+b_{1}$ takes these different values,
the positive integer $k$ has distinct values.
The number of coprime unordered pairs $\left(a_{1}, b_{1}\right)$ for $a_{1}+b_{1}=k$ (where $k$ is odd) is $\frac{\varphi(k)}{2}$ (where $\varphi(k)$ is the Euler's totient function). Therefore, the value of $k$ is equal to the number of coprime unordered pairs $\left(a_{1}, b_{1}\right)$, which is
$$
\begin{array}{l}
=1007 . \\
\end{array}
$$