One, (20 points) Given that $n$ and $k$ are positive integers, and satisfy the inequality
$$
\frac{1}{7}<\frac{n-k}{n+k}<\frac{63}{439} .
$$

If for a given positive integer $n$, there is only one positive integer $k$ that makes the inequality true. Find the maximum and minimum values of all positive integers $n$ that meet the requirement.
From the known inequality, we have
$$
\frac{188}{251} n < k < \frac{189}{251} n.
$$
When $n=252, 253, 254$, substituting into equation (1) respectively, we get
$$
\begin{array}{l}
188 \frac{188}{251} < k < 189, \\
189 \frac{125}{251} < k < 189 \frac{3}{4}, \\
190 \frac{62}{251} < k < 190 \frac{1}{2},
\end{array}
$$

none of which meet the requirements.
When $n=255$, $190 \frac{250}{251} < k < 191 \frac{1}{4}$.
Thus, there is a unique positive integer $k=191$.
Therefore, $n=255$ is the smallest value that meets the condition.
Hence, the maximum value of all positive integers $n$ that meet the requirement is 2008, and the minimum value is 191.