For a positive integer $n>1$, let $g(n)$ denote the largest positive proper divisor of $n$ and $f(n)=n-g(n)$. For example, $g(10)=5, f(10)=5$ and $g(13)=1,f(13)=12$. Let $N$ be the smallest positive integer such that $f(f(f(N)))=97$. Find the largest integer not exceeding $\sqrt{N}$
1. **Define the functions and initial conditions:**
   - For a positive integer \( n > 1 \), let \( g(n) \) denote the largest positive proper divisor of \( n \).
   - Define \( f(n) = n - g(n) \).
   - We need to find the smallest positive integer \( N \) such that \( f(f(f(N))) = 97 \).

2. **First iteration:**
   - Let \( f^2(N) \) be the result after applying \( f \) twice.
   - Let \( p_1 \) be the smallest prime divisor of \( f^2(N) \).
   - By definition, \( f^2(N) - \frac{f^2(N)}{p_1} = 97 \).
   - Solving for \( f^2(N) \):
     \[
     f^2(N) - \frac{f^2(N)}{p_1} = 97 \implies f^2(N) \left(1 - \frac{1}{p_1}\right) = 97 \implies f^2(N) = \frac{97p_1}{p_1 - 1}
     \]
   - Since \( \gcd(p_1 - 1, p_1) = 1 \), the smallest prime \( p_1 \) must be 2.
   - Therefore, \( f^2(N) = \frac{97 \cdot 2}{2 - 1} = 194 \).

3. **Second iteration:**
   - Let \( p_2 \) be the smallest prime divisor of \( f(N) \).
   - By definition, \( f(N) - \frac{f(N)}{p_2} = 194 \).
   - Solving for \( f(N) \):
     \[
     f(N) - \frac{f(N)}{p_2} = 194 \implies f(N) \left(1 - \frac{1}{p_2}\right) = 194 \implies f(N) = \frac{194p_2}{p_2 - 1}
     \]
   - The smallest prime \( p_2 \) can be 2 or 3. We test both:
     - For \( p_2 = 2 \):
       \[
       f(N) = \frac{194 \cdot 2}{2 - 1} = 388
       \]
     - For \( p_2 = 3 \):
       \[
       f(N) = \frac{194 \cdot 3}{3 - 1} = 291
       \]
     - We choose \( p_2 = 2 \) to minimize \( N \), so \( f(N) = 388 \).

4. **Third iteration:**
   - Let \( p_3 \) be the smallest prime divisor of \( N \).
   - By definition, \( N - \frac{N}{p_3} = 388 \).
   - Solving for \( N \):
     \[
     N - \frac{N}{p_3} = 388 \implies N \left(1 - \frac{1}{p_3}\right) = 388 \implies N = \frac{388p_3}{p_3 - 1}
     \]
   - The smallest prime \( p_3 \) can be 2, 3, or 389. We test:
     - For \( p_3 = 2 \):
       \[
       N = \frac{388 \cdot 2}{2 - 1} = 776
       \]
     - For \( p_3 = 3 \):
       \[
       N = \frac{388 \cdot 3}{3 - 1} = 582
       \]
     - For \( p_3 = 389 \):
       \[
       N = \frac{388 \cdot 389}{389 - 1} = 389
       \]
     - We choose \( p_3 = 389 \) to minimize \( N \), so \( N = 389 \).

5. **Final calculation:**
   - We need to find the largest integer not exceeding \( \sqrt{N} \):
     \[
     \sqrt{389} \approx 19.72 \implies \lfloor \sqrt{389} \rfloor = 19
     \]

The final answer is \( \boxed{19} \)