1. We need the smallest equivalence relation $S$ on $A=\{1,2,3,4\}$ such that $R=\{(1,2),(2,3),(1,4)\}\subset S.$

An equivalence relation must be:

• reflexive,
• symmetric,
• transitive.

So we must add the minimum number of ordered pairs to make $R$ satisfy all three properties.

2. Start with the given pairs: $(1,2),\ (2,3),\ (1,4).$

These show that:

• $1$ is related to $2$,
• $2$ is related to $3$,
• $1$ is related to $4$.

Since an equivalence relation is symmetric, we must also have: $(2,1),\ (3,2),\ (4,1).$

3. Now use transitivity.

Because $(1,2) \in S \quad \text{and} \quad (2,3) \in S,$ transitivity gives $(1,3) \in S.$

Then by symmetry, $(3,1) \in S.$

Also, from $(2,1) \in S \quad \text{and} \quad (1,4) \in S,$ we get $(2,4) \in S,$ and by symmetry, $(4,2) \in S.$

Similarly, from $(3,2) \in S \quad \text{and} \quad (2,1) \in S,$ we get $(3,1)$, already included, and with $(1,4)$ we get $(3,4) \in S,$ and by symmetry, $(4,3) \in S.$

Thus all elements $1,2,3,4$ become mutually related. So they all lie in the same equivalence class.

4. In any equivalence relation, every element must be related to itself, so reflexivity gives: $(1,1),(2,2),(3,3),(4,4).$

Since all 4 elements are in one equivalence class, the equivalence relation is simply $A \times A.$

Therefore, $|S|=|A\times A|=4\times 4=16.$

5. Hence the minimum value of $n$ is $\boxed{16}.$