For all positive integers $n > 1$, let $f(n)$ denote the largest odd proper divisor of $n$ (a proper divisor of $n$ is a positive divisor of $n$ except for $n$ itself). Given that $N=20^{23}\cdot23^{20}$, compute
\[\frac{f(N)}{f(f(f(N)))}.\]
1. **Identify the largest odd proper divisor of \( N \):**
   Given \( N = 20^{23} \cdot 23^{20} \), we first express 20 in terms of its prime factors:
   \[
   20 = 2^2 \cdot 5
   \]
   Therefore,
   \[
   N = (2^2 \cdot 5)^{23} \cdot 23^{20} = 2^{46} \cdot 5^{23} \cdot 23^{20}
   \]
   The largest odd proper divisor of \( N \) is obtained by removing all factors of 2 (since 2 is the only even prime factor):
   \[
   f(N) = 5^{23} \cdot 23^{20}
   \]

2. **Compute \( f(f(N)) \):**
   Now, we need to find the largest odd proper divisor of \( f(N) = 5^{23} \cdot 23^{20} \). Since \( f(N) \) is already odd, we need to consider its proper divisors:
   \[
   f(f(N)) = f(5^{23} \cdot 23^{20})
   \]
   The largest proper divisor of \( 5^{23} \) is \( 5^{22} \), and the largest proper divisor of \( 23^{20} \) is \( 23^{19} \). However, since we are looking for the largest odd proper divisor, we only need to reduce the power of one of the factors:
   \[
   f(f(N)) = 5^{22} \cdot 23^{20}
   \]

3. **Compute \( f(f(f(N))) \):**
   Next, we find the largest odd proper divisor of \( f(f(N)) = 5^{22} \cdot 23^{20} \):
   \[
   f(f(f(N))) = f(5^{22} \cdot 23^{20})
   \]
   Similarly, the largest proper divisor of \( 5^{22} \) is \( 5^{21} \), and the largest proper divisor of \( 23^{20} \) is \( 23^{19} \). Again, we only need to reduce the power of one of the factors:
   \[
   f(f(f(N))) = 5^{21} \cdot 23^{20}
   \]

4. **Compute the ratio \( \frac{f(N)}{f(f(f(N)))} \):**
   Finally, we compute the ratio:
   \[
   \frac{f(N)}{f(f(f(N)))} = \frac{5^{23} \cdot 23^{20}}{5^{21} \cdot 23^{20}} = \frac{5^{23}}{5^{21}} = 5^{23-21} = 5^2 = 25
   \]

Conclusion:
\[
\boxed{25}
\]