Given the problem, we need to determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the number of divisors of $sn$ and $sk$ are equal.

To solve this problem, we use the property that the number of divisors $d(x)$ of an integer $x$ is determined by its prime factorization. Suppose $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$ and $k = q_1^{b_1} q_2^{b_2} \cdots q_t^{b_t}$ where $p_i$ and $q_i$ are primes.

$$d(sn) = d(s) \cdot d(n) \quad \text{and} \quad d(sk) = d(s) \cdot d(k).$$

Given $d(sn) = d(sk)$, we have:

$$d(n) = d(k).$$

This means that $n$ and $k$ must have the same divisor count. If $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$ and $k = q_1^{b_1} q_2^{b_2} \cdots q_t^{b_t}$, then:

$$(a_1 + 1)(a_2 + 1) \cdots (a_r + 1) = (b_1 + 1)(b_2 + 1) \cdots (b_t + 1).$$

For the pair $(n, k)$ to satisfy $d(n) = d(k)$ with a valid $s$, $n$ and $k$ should not be related as divisibility by each other; otherwise, one would directly have a greater count of divisors through multiplication by any factor $s$ that includes extra prime factors from $n$ or $k$.

Thus, a necessary condition is that neither integer divides the other, ensuring complete freedom in choosing $s$ to balance out the divisor counts.

Therefore, all pairs $(m, n)$ satisfying the conditions are those for which:
$$m \nmid n \quad \text{and} \quad n \nmid m.$$

The solution is given by:
$$\boxed{\text{all pairs } (m, n) \text{ such that } m \nmid n, n \nmid m.}$$