For each positive integer $n$, consider the highest common factor $h_n$ of the two numbers $n!+1$ and $(n+1)!$. For $n<100$, find the largest value of $h_n$.
1. **Define the problem and the greatest common divisor (gcd):**
   We need to find the highest common factor \( h_n \) of the two numbers \( n! + 1 \) and \( (n+1)! \) for \( n < 100 \). This can be expressed as:
   \[
   h_n = \gcd(n! + 1, (n+1)!)
   \]

2. **Express \((n+1)!\) in terms of \(n!\):**
   \[
   (n+1)! = (n+1) \cdot n!
   \]

3. **Use the properties of gcd:**
   Since \( h_n \) divides both \( n! + 1 \) and \((n+1)! \), we have:
   \[
   h_n \mid (n! + 1) \quad \text{and} \quad h_n \mid (n+1) \cdot n!
   \]

4. **Simplify the gcd expression:**
   Since \( h_n \mid (n+1) \cdot n! \) and \( h_n \mid n! + 1 \), we can write:
   \[
   h_n \mid (n+1) \cdot n! + n! = (n+1) \cdot n! + n! = (n+1+1) \cdot n! = (n+1) \cdot n!
   \]
   Therefore, \( h_n \mid n! + 1 \) and \( h_n \mid (n+1) \cdot n! \).

5. **Further simplification:**
   \[
   h_n \mid (n+1) \cdot n! + n! - (n+1) \cdot n! = n!
   \]
   Thus, \( h_n \mid n! + 1 - n! = 1 \).

6. **Case analysis:**
   - **Case 1:** If \( h_n = n+1 \), then \( n+1 \) must divide \( n! + 1 \). This implies:
     \[
     n! \equiv -1 \pmod{n+1}
     \]
     By Wilson's Theorem, this is true if and only if \( n+1 \) is a prime number.
   - **Case 2:** If \( h_n < n+1 \), then \( h_n \leq n \). Since \( h_n \mid n! \) and \( h_n \mid n! + 1 \), we have:
     \[
     h_n \mid (n! + 1) - n! = 1
     \]
     Therefore, \( h_n = 1 \).

7. **Conclusion from the cases:**
   - If \( n+1 \) is a prime number, then \( h_n = n+1 \).
   - If \( n+1 \) is not a prime number, then \( h_n = 1 \).

8. **Finding the largest value of \( h_n \) for \( n < 100 \):**
   The largest value of \( h_n \) occurs when \( n+1 \) is the largest prime number less than 100. The largest prime number less than 100 is 97. Therefore, \( n = 96 \).

The final answer is \(\boxed{97}\)