Let $n$ be a fixed positive integer and let $x, y$ be positive integers such that $xy = nx+ny$. 
Determine the minimum and the maximum of  $x$ in terms of $n$.
1. Start with the given equation:
   \[
   xy = nx + ny
   \]
   Rearrange the equation to isolate \(x\):
   \[
   xy - nx = ny
   \]
   Factor out \(x\) on the left-hand side:
   \[
   x(y - n) = ny
   \]
   Solve for \(x\):
   \[
   x = \frac{ny}{y - n}
   \]

2. Since \(x\) and \(y\) are positive integers, and \(x > 0\), the denominator \(y - n\) must be positive. Therefore, we have:
   \[
   y > n \implies y \geq n + 1
   \]

3. To find the minimum value of \(x\), consider the smallest possible value of \(y\), which is \(y = n + 1\):
   \[
   x = \frac{n(n + 1)}{(n + 1) - n} = \frac{n(n + 1)}{1} = n(n + 1)
   \]
   Thus, the minimum value of \(x\) is:
   \[
   x_{\text{min}} = n + 1
   \]

4. To find the maximum value of \(x\), consider the largest possible value of \(y\). Since \(y\) can be arbitrarily large, we need to analyze the behavior of \(x\) as \(y\) increases. As \(y\) approaches infinity, the fraction \(\frac{ny}{y - n}\) approaches \(n\):
   \[
   \lim_{y \to \infty} \frac{ny}{y - n} = n
   \]
   However, since \(x\) must be an integer, the maximum value of \(x\) is achieved when \(y\) is just large enough to make \(x\) an integer. This is when \(y = n + 1\):
   \[
   x_{\text{max}} = n(n + 1)
   \]

5. Therefore, the range of \(x\) is:
   \[
   n + 1 \leq x \leq n(n + 1)
   \]

6. Both bounds are achievable. For \(x = n + 1\), we can set \(y = n(n + 1)\):
   \[
   (n + 1) \cdot n(n + 1) = n(n + 1) + n(n + 1)
   \]
   For \(x = n(n + 1)\), we can set \(y = n + 1\):
   \[
   n(n + 1) \cdot (n + 1) = n(n + 1) + n(n + 1)
   \]

The final answer is \( \boxed{ n + 1 \leq x \leq n(n + 1) } \)