For any positive integer $n$ , let $s(n)$ denote the number of ordered pairs $(x,y)$ of positive integers for which $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$ . Determine the set of positive integers for which $s(n) = 5$
1. We start with the given equation:
   \[
   \frac{1}{x} + \frac{1}{y} = \frac{1}{n}
   \]
   By finding a common denominator, we can rewrite this as:
   \[
   \frac{x + y}{xy} = \frac{1}{n}
   \]
   Multiplying both sides by \( xy \cdot n \) gives:
   \[
   n(x + y) = xy
   \]
   Rearranging terms, we get:
   \[
   xy - nx - ny = 0
   \]
   Adding \( n^2 \) to both sides, we can factor the left-hand side:
   \[
   xy - nx - ny + n^2 = n^2 \implies (x - n)(y - n) = n^2
   \]

2. The equation \((x - n)(y - n) = n^2\) implies that \(x - n\) and \(y - n\) are pairs of divisors of \(n^2\). Let \(d\) be a divisor of \(n^2\), then we can write:
   \[
   x - n = d \quad \text{and} \quad y - n = \frac{n^2}{d}
   \]
   Thus:
   \[
   x = d + n \quad \text{and} \quad y = \frac{n^2}{d} + n
   \]

3. We need to find the number of ordered pairs \((x, y)\) such that \(x\) and \(y\) are positive integers. This is equivalent to finding the number of divisors of \(n^2\). Let \(n\) have the prime factorization:
   \[
   n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}
   \]
   Then the number of divisors of \(n^2\) is given by:
   \[
   \sigma(n^2) = (2e_1 + 1)(2e_2 + 1) \cdots (2e_k + 1)
   \]

4. We are given that \(s(n) = 5\). This means that \(n^2\) must have exactly 5 divisors. The number 5 is a prime number, and the only way to get 5 as a product of integers of the form \((2e_i + 1)\) is if one of the factors is 5 and the rest are 1. This implies:
   \[
   2e_1 + 1 = 5 \quad \text{and} \quad 2e_2 + 1 = 1, \; 2e_3 + 1 = 1, \; \ldots
   \]
   Solving \(2e_1 + 1 = 5\), we get:
   \[
   e_1 = 2
   \]
   For the other exponents, \(2e_i + 1 = 1\) implies \(e_i = 0\).

5. Therefore, \(n\) must be of the form \(p^2\) where \(p\) is a prime number. The set of such \(n\) is:
   \[
   \{p^2 \mid p \text{ is a prime number}\}
   \]

The final answer is \( \boxed{ \{p^2 \mid p \text{ is a prime number}\} } \).