6. Note that $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ and $\frac{1}{6}-\frac{1}{7}=\frac{1}{6 \times 7}$ and $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=1$.
Determine one triple $(x, y, z)$ of positive integers with $1000<x<y<z<2000$ and
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{45}=1
$$
6. The given fact that $\frac{1}{6}-\frac{1}{7}=\frac{1}{6 \times 7}$ suggests that we consider $\frac{1}{n}-\frac{1}{n+1}$.
Using a common denominator,
$$
\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n(n+1)}
$$

Starting with
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=1
$$
we subtract and add the same quantities from the left side to obtain
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}-\frac{1}{43}+\frac{1}{43}-\frac{1}{44}+\frac{1}{44}-\frac{1}{45}+\frac{1}{45}=1
$$

We then regroup the terms on the left side as
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\left(\frac{1}{42}-\frac{1}{43}\right)+\left(\frac{1}{43}-\frac{1}{44}\right)+\left(\frac{1}{44}-\frac{1}{45}\right)+\frac{1}{45}=1
$$
and use the fact $(*)$ above to obtain
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42 \times 43}+\frac{1}{43 \times 44}+\frac{1}{44 \times 45}+\frac{1}{45}=1
$$

This gives
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{1806}+\frac{1}{1892}+\frac{1}{1980}+\frac{1}{45}=1
$$

Therefore, a solution to the given equation that satisfies the condition $1000<x<y<z<2000$ is $(x, y, z)=(1806,1892,1980)$.
There are other solutions to this equation obeying the given conditions, but we are only asked to find one such solution.
ANSWER: $(1806,1892,1980)$