Let $n$ be the number of mathematicians at the conference. We are tasked with finding the smallest value of $k$ such that there are at least three mathematicians, say $X, Y,$ and $Z$, who are all acquainted with each other (i.e., each knows the other two).

We can model this problem using graph theory, where each mathematician is represented by a vertex, and an edge exists between two vertices if the corresponding mathematicians know each other. Our goal is to find the smallest $k$ such that any graph with $n$ vertices, each with degree $k$, must contain a triangle (a cycle of three vertices).

To approach this problem, consider the following:

1. The total number of acquaintances (edges) in the graph is $\frac{nk}{2}$, since each of the $n$ vertices has degree $k$.

2. The problem is equivalent to finding the smallest $k$ such that every subgraph $G$ of $n$ vertices and $\frac{nk}{2}$ edges contains at least one triangle.

Let's consider a complete bipartite graph $K_{\lfloor \frac{n}{2} \rfloor, \lceil \frac{n}{2} \rceil}$. This graph divides the $n$ vertices into two sets of sizes $\lfloor \frac{n}{2} \rfloor$ and $\lceil \frac{n}{2} \rceil$, and every vertex in one set is connected to every vertex in the other set, with no edges within the same set. The degree of each vertex in this graph is $\lfloor \frac{n}{2} \rfloor$. Such a graph contains no triangles as there are no edges within the same set.

If $k \leq \lfloor \frac{n}{2} \rfloor$, it's possible to arrange the acquaintance relations as in a complete bipartite graph, avoiding triangles. To ensure a triangle always exists, we need at least $k > \lfloor \frac{n}{2} \rfloor$.

Thus, adding one more acquaintance to each mathematician ensures that at least one triangle must form due to the increase in connectivity. Therefore, the smallest $k$ for which a triangle always exists is:

$$k = \left\lfloor \frac{n}{2} \right\rfloor + 1.$$

Thus, the required smallest value of $k$ is:

$$\boxed{\left\lfloor \frac{n}{2} \right\rfloor + 1}.$$