For which integers $n>1$ does there exist a rectangle that can be subdivided into $n$ pairwise noncongruent rectangles similar to the original rectangle?
To determine for which integers \( n > 1 \) there exists a rectangle that can be subdivided into \( n \) pairwise noncongruent rectangles similar to the original rectangle, we proceed as follows:

1. **Case \( n = 2 \)**:
   - If \( n = 2 \), we need to subdivide the original rectangle into two smaller rectangles that are similar to the original and noncongruent to each other. This is impossible because if two rectangles are similar to the original and noncongruent, their aspect ratios must be the same as the original, which contradicts the requirement of noncongruence. Therefore, \( n = 2 \) is not possible.

2. **Case \( n \ge 3 \)**:
   - For \( n \ge 3 \), we can construct a spiral of rectangles, each touching two of the previous rectangles in alternating orientations, except for the \( n \)-th rectangle, which we orient in the same way as the \((n-1)\)-st rectangle. This construction ensures that all rectangles are pairwise noncongruent and similar to the original rectangle.

3. **Aspect Ratio Analysis**:
   - Let \( r > 1 \) be a real number to be determined. We need to ensure that the aspect ratio \( a(r) \) of the entire large rectangle matches \( r \). 
   - When \( r = 1 \), we have a golden spiral, and the aspect ratio \( a(1) \) is given by the ratio of consecutive Fibonacci numbers, \( \frac{F_{k+1}}{F_k} \), which is greater than 1.
   - As \( r \to \infty \), the aspect ratio \( a(r) \) approaches a rational polynomial in \( r \), specifically \( \left( \frac{1}{2} + o(1) \right) r \), which is less than \( r \).

4. **Intermediate Value Theorem**:
   - By the Intermediate Value Theorem, since \( a(1) > 1 \) and \( a(r) < r \) as \( r \to \infty \), there must exist some \( r \) such that \( a(r) = r \). This ensures that the constructed rectangles are similar to the original rectangle.

Therefore, for \( n \ge 3 \), it is possible to subdivide a rectangle into \( n \) pairwise noncongruent rectangles similar to the original rectangle.

The final answer is \( \boxed{ n \ge 3 } \).