1. Graph Representation and Vertex Coloring:
- Consider a graph where each vertex represents one of the $2^7 = 128$ possible weekly schedules for an employee (each day can either be a workday or a non-workday).
- Connect two vertices with an edge if their corresponding schedules differ by exactly one day. This means each vertex has a degree of 7.

2. Vertex Coloring:
- Let $v_1, v_2, \ldots, v_n$ be the vertices representing the weekly schedules of the $n$ employees.
- Paint each vertex $v_i$ black.
- Paint each of the 7 neighbors of each $v_i$ white.

3. Properties of the Coloring:
- No two black vertices are adjacent because if they were, their schedules would differ by exactly one day, contradicting the condition that any two employees' schedules differ by at least 3 days.
- No vertex is painted white more than once. If a vertex $v$ were painted white more than once, it would mean $v$ is a neighbor of both $v_i$ and $v_j$. This implies $v_i$ and $v_j$ differ by at most 2 days, which contradicts the given condition.

4. Counting Vertices:
- Each black vertex has 7 white neighbors, and there are $n$ black vertices.
- Thus, the total number of painted vertices is $8n$ (since each black vertex and its 7 neighbors are distinct).
- Since there are only 128 vertices in total, we have:
$$8n \leq 128 \implies n \leq 16$$

5. Constructing an Example with 16 Employees:
- We need to show that it is possible to have 16 employees with schedules satisfying the given conditions.
- Consider the following 16 weekly schedules represented as 7-tuples of 0s and 1s:
$$\begin{aligned} & (0,0,0,0,0,0,0), (1,1,1,0,0,0,0), (1,0,0,1,1,0,0), (1,0,0,0,0,1,1), \\ & (0,1,0,1,0,0,1), (0,1,0,0,1,1,0), (0,0,1,1,0,1,0), (0,0,1,0,1,0,1), \\ & (1,1,1,1,1,1,1), (0,0,0,1,1,1,1), (0,1,1,0,0,1,1), (0,1,1,1,1,0,0), \\ & (1,0,1,0,1,1,0), (1,0,1,1,0,0,1), (1,1,0,0,1,0,1), (1,1,0,1,0,1,0) \end{aligned}$$
- Verify that for any two of these schedules, there are at least 3 days where one employee works and the other does not. This can be checked by comparing each pair of schedules and ensuring they differ in at least 3 positions.

The final answer is $\boxed{ n \leq 16 }$