We need to determine whether there exist positive integers $a$ and $b$ such that $a$ does not divide $b^n - n$ for all positive integers $n$.

To solve this, we claim that for each pair of positive integers $(a, b)$, there exists some $n$ such that $a \mid b^n - n$. Consider the infinite sequence of positive integers defined as $b_1 = b$ and $b_{k+1} = b^{b_k}$. We will use the following lemma:

Lemma (USAMO 1991 P3): The sequence $\{b_k\}_{k \geq 1}$ eventually becomes constant modulo $a$.

Proof of Lemma: We use strong induction on $a$. For $a = 1, 2$, the result is obvious. Suppose our claim is true for $1, 2, \ldots, a-1$. Consider the case when $\gcd(a, b) = d > 1$. Choose some prime divisor $p$ of $d$. Note that, for sufficiently large $j$, $p^{\nu_p(a)} \mid b_j$. So we can effectively ignore $d$, and assume that $\gcd(a, b) = 1$. Then it is well known that
$$b_{k+1} = b^{b_k} \equiv b^{b_k \pmod{\phi(a)}} \pmod{a}.$$
By the induction hypothesis, for sufficiently large $k$, the sequence $\{b_k\}_{k \geq 1}$ eventually becomes constant modulo $\phi(a)$, giving the desired result. $\Box$

Returning to the problem, by choosing $n = b_k$ for sufficiently large $k$, we conclude that $a \mid b^n - n$. Hence, there do not exist positive integers $a$ and $b$ such that $a$ does not divide $b^n - n$ for all positive integers $n$.

The answer is: \boxed{\text{No}}.