The sequence of positive integers $a_0, a_1, a_2, . . .$ is defined by $a_0 = 3$ and $$a_{n+1} - a_n = n(a_n - 1)$$ for all $n \ge 0$. Determine all integers $m \ge 2$ for which $gcd (m, a_n) = 1$ for all $n \ge 0$.
1. **Rearrange the given recursion:**
   The given recursion is:
   \[
   a_{n+1} - a_n = n(a_n - 1)
   \]
   Rearrange it to:
   \[
   a_{n+1} - 1 = (n+1)(a_n - 1)
   \]

2. **Telescoping argument:**
   We can use a telescoping argument to find a general form for \(a_n\). Start with \(a_0 = 3\):
   \[
   a_0 - 1 = 2
   \]
   For \(n = 0\):
   \[
   a_1 - 1 = 1(a_0 - 1) = 1 \cdot 2 = 2 \implies a_1 = 3
   \]
   For \(n = 1\):
   \[
   a_2 - 1 = 2(a_1 - 1) = 2 \cdot 2 = 4 \implies a_2 = 5
   \]
   For \(n = 2\):
   \[
   a_3 - 1 = 3(a_2 - 1) = 3 \cdot 4 = 12 \implies a_3 = 13
   \]
   We observe that:
   \[
   a_n - 1 = 2 \cdot n!
   \]
   Therefore:
   \[
   a_n = 2 \cdot n! + 1
   \]

3. **Check the gcd condition:**
   We need to determine all integers \(m \ge 2\) for which \(\gcd(m, a_n) = 1\) for all \(n \ge 0\). Since \(a_n = 2 \cdot n! + 1\), we note that \(a_n\) is always odd. Therefore, \(m\) must be odd to ensure \(\gcd(m, a_n) = 1\).

4. **Check for primes \(p \ge 3\):**
   We need to show that for any prime \(p \ge 3\), there exists an \(n\) such that \(p \mid a_n\). Consider \(p = 3\):
   \[
   a_0 = 3 \implies 3 \mid a_0
   \]
   For \(p \ge 5\), by Wilson's theorem, \((p-1)! \equiv -1 \pmod{p}\). Therefore:
   \[
   (p-3)! \equiv \frac{-1}{(p-2)(p-1)} \equiv \frac{-1}{2 \cdot (p-1)} \equiv -\frac{1}{2} \pmod{p}
   \]
   Consequently:
   \[
   a_{p-3} = 2 \cdot (p-3)! + 1 \equiv 2 \cdot \left(-\frac{1}{2}\right) + 1 \equiv -1 + 1 \equiv 0 \pmod{p}
   \]
   Thus, \(p \mid a_{p-3}\).

5. **Conclusion:**
   Since \(a_n\) is odd for all \(n\) and for any prime \(p \ge 3\), there exists an \(n\) such that \(p \mid a_n\), the only possible values for \(m\) are powers of 2.

The final answer is \( \boxed{ m = 2^k } \), where \(k \ge 1\).