6. Given that $f(x)$ is an odd function defined on $\mathbf{R}$, $f(1)=1$, and for any $x<0$, we have
$$
\begin{array}{l}
f\left(\frac{1}{x}\right)=x f\left(\frac{1}{1-x}\right) . \\
\text { Then } \sum_{k=1}^{1009} f\left(\frac{1}{k}\right) f\left(\frac{1}{2019-k}\right)=
\end{array}
$$
$\qquad$
6. $\frac{2^{2016}}{2017!}$.

Let $x=-k\left(k \in \mathbf{Z}_{+}\right)$. Substituting into the functional equation, we get
$$
\begin{array}{l}
f\left(-\frac{1}{k}\right)=-k f\left(\frac{1}{1+k}\right) \\
\Rightarrow f\left(\frac{1}{k}\right)=k f\left(\frac{1}{1+k}\right) \\
\Rightarrow \frac{f\left(\frac{1}{1+k}\right)}{f\left(\frac{1}{k}\right)}=\frac{1}{k} .
\end{array}
$$

Thus, $f\left(\frac{1}{k}\right)=\frac{1}{(k-1)!}$.
Then the original expression $=\sum_{k=1}^{1009} \frac{1}{(k-1)!\cdot(2018-k)!}$
$$
\begin{array}{l}
=\sum_{k=0}^{1008} \frac{1}{k!\cdot(2017-k)!} \\
=\frac{1}{2017!} \sum_{k=0}^{1008} \mathrm{C}_{2017}^{k}=\frac{2^{2016}}{2017!} .
\end{array}
$$