An integer $n \geq 3$ is called [i]special[/i] if it does not divide $\left ( n-1 \right )!\left ( 1+\frac{1}{2}+\cdot \cdot \cdot +\frac{1}{n-1} \right )$. Find all special numbers $n$ such that $10 \leq n \leq 100$.
To find all special numbers \( n \) such that \( 10 \leq n \leq 100 \), we need to determine which integers \( n \) do not divide the expression \((n-1)!\left(1 + \frac{1}{2} + \cdots + \frac{1}{n-1}\right)\).

1. **Case 1: \( n = p \) for some prime \( p \)**

   - For \( n = 2 \):
     \[
     (n-1)!\left(1 + \frac{1}{2} + \cdots + \frac{1}{n-1}\right) = 1
     \]
     which is not divisible by \( 2 \), so \( 2 \) is special.

   - For \( n = p \) where \( p \) is an odd prime:
     \[
     1 + \frac{1}{2} + \cdots + \frac{1}{p-1} \equiv 0 \pmod{p}
     \]
     by Vieta's formulas in the field \(\mathbb{F}_p\). Thus, \( n \) is not special for odd primes \( p \).

2. **Case 2: \( n = p^k \) for some prime \( p \) and integer \( k \geq 2 \)**

   - For \( n = 4 \):
     \[
     (n-1)!\left(1 + \frac{1}{2} + \cdots + \frac{1}{n-1}\right) = 6 \left(1 + \frac{1}{2} + \frac{1}{3}\right) = 6 \cdot \frac{11}{6} = 11
     \]
     which is not divisible by \( 4 \), so \( 4 \) is special.

   - For \( n = 8 \):
     \[
     (n-1)!\left(1 + \frac{1}{2} + \cdots + \frac{1}{n-1}\right) = 5040 \left(1 + \frac{1}{2} + \cdots + \frac{1}{7}\right) = 5040 \cdot \frac{363}{140} = 13068
     \]
     which is not divisible by \( 8 \), so \( 8 \) is special.

   - For \( n = 9 \):
     \[
     (n-1)!\left(1 + \frac{1}{2} + \cdots + \frac{1}{n-1}\right) = 40320 \left(1 + \frac{1}{2} + \cdots + \frac{1}{8}\right) = 40320 \cdot \frac{761}{280} = 109584
     \]
     which is divisible by \( 9 \), so \( 9 \) is not special.

   - For \( n = p^k \) with \( k \geq 2 \) and \( n \neq 4, 8, 9 \):
     \[
     v_p((p^k-1)!) - v_p(m) - 1 \geq k
     \]
     This inequality holds for \( p \geq 5 \) when \( k = 2 \), \( p \geq 3 \) when \( k = 3 \), and for all \( p \) when \( k \geq 4 \). Thus, \( n \) is not special for these values.

3. **Case 3: \( n = 2p \) for some odd prime \( p \)**

   - For \( n = 2p \):
     \[
     (n-1)!\left(1 + \frac{1}{2} + \cdots + \frac{1}{n-1}\right) \equiv 1 \pmod{p}
     \]
     This implies that the sum is \( 1 \pmod{p} \), so it is not divisible by \( 2p \). Thus, all \( n = 2p \) with \( p \) an odd prime are special.

4. **Case 4: \( n \) is not a perfect prime power and \( n \) is not \( 2p \) for some odd prime \( p \)**

   - For such \( n \):
     \[
     n \mid (n-1)!\left(1 + \frac{1}{2} + \cdots + \frac{1}{n-1}\right)
     \]
     This implies \( n \) is not special.

Combining all cases, the special numbers \( n \in [10, 100] \) are \( 10, 14, 22, 26, 34, 38, 46, 58, 62, 74, 82, 86, 94 \).

The final answer is \( \boxed{ \{10, 14, 22, 26, 34, 38, 46, 58, 62, 74, 82, 86, 94\} } \).