Find all positive integer $n$ such that
$$\frac{\sin{n\theta}}{\sin{\theta}} - \frac{\cos{n\theta}}{\cos{\theta}} = n-1$$
holds for all $\theta$ which are not integral multiples of $\frac{\pi}{2}$ 
1. Start with the given equation:
   \[
   \frac{\sin{n\theta}}{\sin{\theta}} - \frac{\cos{n\theta}}{\cos{\theta}} = n-1
   \]

2. Combine the fractions on the left-hand side:
   \[
   \frac{\sin{n\theta} \cos{\theta} - \cos{n\theta} \sin{\theta}}{\sin{\theta} \cos{\theta}} = n-1
   \]

3. Use the sine addition formula, \(\sin{(a - b)} = \sin{a} \cos{b} - \cos{a} \sin{b}\):
   \[
   \frac{\sin{((n-1)\theta)}}{\sin{\theta} \cos{\theta}} = n-1
   \]

4. Multiply both sides by \(\sin{\theta} \cos{\theta}\):
   \[
   \sin{((n-1)\theta)} = (n-1) \sin{\theta} \cos{\theta}
   \]

5. Use the double-angle identity, \(\sin{2\theta} = 2 \sin{\theta} \cos{\theta}\):
   \[
   \sin{((n-1)\theta)} = \frac{n-1}{2} \sin{2\theta}
   \]

6. Let \(\theta = \frac{\pi}{4}\):
   \[
   \sin{\left((n-1) \frac{\pi}{4}\right)} = \frac{n-1}{2}
   \]

7. The sine function can take values in the range \([-1, 1]\). Therefore, \(\frac{n-1}{2}\) must be within this range:
   \[
   -1 \leq \frac{n-1}{2} \leq 1
   \]

8. Solve the inequality:
   \[
   -1 \leq \frac{n-1}{2} \leq 1
   \]
   \[
   -2 \leq n-1 \leq 2
   \]
   \[
   -1 \leq n \leq 3
   \]

9. Since \(n\) is a positive integer, the possible values for \(n\) are:
   \[
   n = 1, 2, 3
   \]

10. Verify each value of \(n\):
    - For \(n = 1\):
      \[
      \frac{\sin{\theta}}{\sin{\theta}} - \frac{\cos{\theta}}{\cos{\theta}} = 0
      \]
      \[
      1 - 1 = 0
      \]
      This holds true.

    - For \(n = 2\):
      \[
      \frac{\sin{2\theta}}{\sin{\theta}} - \frac{\cos{2\theta}}{\cos{\theta}} = 1
      \]
      \[
      2 \cos{\theta} - (2 \cos^2{\theta} - 1) = 1
      \]
      \[
      2 \cos{\theta} - 2 \cos^2{\theta} + 1 = 1
      \]
      \[
      2 \cos{\theta} - 2 \cos^2{\theta} = 0
      \]
      \[
      2 \cos{\theta} (1 - \cos{\theta}) = 0
      \]
      This holds true for \(\cos{\theta} = 0\), but \(\theta\) cannot be an integral multiple of \(\frac{\pi}{2}\). Therefore, \(n = 2\) is not a valid solution.

    - For \(n = 3\):
      \[
      \frac{\sin{3\theta}}{\sin{\theta}} - \frac{\cos{3\theta}}{\cos{\theta}} = 2
      \]
      \[
      4 \cos^2{\theta} - 1 - (4 \cos^3{\theta} - 3 \cos{\theta}) = 2
      \]
      \[
      4 \cos^2{\theta} - 1 - 4 \cos^3{\theta} + 3 \cos{\theta} = 2
      \]
      \[
      4 \cos^2{\theta} - 4 \cos^3{\theta} + 3 \cos{\theta} - 1 = 2
      \]
      \[
      4 \cos^2{\theta} - 4 \cos^3{\theta} + 3 \cos{\theta} - 3 = 0
      \]
      This holds true for \(\cos{\theta} = 0\), but \(\theta\) cannot be an integral multiple of \(\frac{\pi}{2}\). Therefore, \(n = 3\) is not a valid solution.

The final answer is \( \boxed{ n = 1 } \)