Given that there are $24$ primes between $3$ and $100$, inclusive, what is the number of ordered pairs $(p, a)$ with $p$ prime, $3 \le p < 100$, and $1 \le a < p$ such that $p \mid (a^{p-2} - a)$?
1. **Identify the primes between 3 and 100:**
   We are given that there are 24 primes between 3 and 100, inclusive. Let's denote these primes as \( p_1, p_2, \ldots, p_{24} \).

2. **Condition for ordered pairs \((p, a)\):**
   We need to find the number of ordered pairs \((p, a)\) such that \( p \) is a prime between 3 and 100, and \( 1 \le a < p \) such that \( p \mid (a^{p-2} - a) \).

3. **Using Fermat's Little Theorem:**
   Fermat's Little Theorem states that for any integer \( a \) and a prime \( p \), \( a^{p-1} \equiv 1 \pmod{p} \) if \( p \nmid a \). This can be rewritten as:
   \[
   a^{p-1} \equiv 1 \pmod{p}
   \]
   Since \( p \mid (a^{p-2} - a) \), we can rewrite this as:
   \[
   a^{p-2} \equiv a \pmod{p}
   \]

4. **Simplifying the condition:**
   If \( p \nmid a \), then we can divide both sides by \( a \) (since \( a \neq 0 \pmod{p} \)):
   \[
   a^{p-3} \equiv 1 \pmod{p}
   \]
   By Fermat's Little Theorem, we also have:
   \[
   a^{p-1} \equiv 1 \pmod{p}
   \]
   This implies that \( a^{p-1} - 1 \equiv 0 \pmod{p} \). Therefore, \( p \mid (a^{p-1} - 1) \).

5. **Finding the solutions:**
   The equation \( a^{p-3} \equiv 1 \pmod{p} \) implies that \( a \) must be such that \( a^2 \equiv 1 \pmod{p} \). The solutions to this congruence are \( a \equiv 1 \pmod{p} \) and \( a \equiv -1 \pmod{p} \).

6. **Counting the valid pairs:**
   For each prime \( p \), there are exactly 2 values of \( a \) that satisfy the condition: \( a = 1 \) and \( a = p-1 \). Since there are 24 primes between 3 and 100, the total number of ordered pairs \((p, a)\) is:
   \[
   2 \times 24 = 48
   \]

Conclusion:
\[
\boxed{48}
\]