An equilateral triangle $ABC$ is divided into $100$ congruent equilateral triangles. What is the greatest number of vertices of small triangles that can be chosen so that no two of them lie on a line that is parallel to any of the sides of the triangle $ABC$?
To solve this problem, we need to determine the maximum number of vertices of the small equilateral triangles that can be chosen such that no two of them lie on a line parallel to any of the sides of the large equilateral triangle \(ABC\).

1. **Understanding the Division of the Triangle:**
   - The large equilateral triangle \(ABC\) is divided into 100 smaller congruent equilateral triangles.
   - This implies that the side length of the large triangle is divided into 10 equal segments (since \(10^2 = 100\)).

2. **Coordinate System and Constraints:**
   - We can place the large triangle in a coordinate system where the vertices of the small triangles have integer coordinates.
   - Let the height of each small triangle be 1 unit.
   - Denote the distances from the sides of the large triangle by \(a_i, b_i, c_i\) for the \(i\)-th point.

3. **Sum of Distances:**
   - For any chosen vertex \(i\), the sum of the distances from the three sides of the large triangle is constant:
     \[
     a_i + b_i + c_i = 10
     \]
   - If we choose 8 vertices, we have:
     \[
     \sum_{i=1}^8 a_i + \sum_{i=1}^8 b_i + \sum_{i=1}^8 c_i = 80
     \]

4. **Minimum Sum of Distances:**
   - Each of the sums \(\sum_{i=1}^8 a_i\), \(\sum_{i=1}^8 b_i\), and \(\sum_{i=1}^8 c_i\) must be at least the sum of the first 8 non-negative integers:
     \[
     0 + 1 + 2 + \ldots + 7 = 28
     \]
   - Therefore:
     \[
     \sum_{i=1}^8 a_i \geq 28, \quad \sum_{i=1}^8 b_i \geq 28, \quad \sum_{i=1}^8 c_i \geq 28
     \]
   - Adding these inequalities, we get:
     \[
     \sum_{i=1}^8 a_i + \sum_{i=1}^8 b_i + \sum_{i=1}^8 c_i \geq 3 \times 28 = 84
     \]
   - This is a contradiction because we previously established that the sum is 80.

5. **Conclusion:**
   - Since choosing 8 vertices leads to a contradiction, the maximum number of vertices that can be chosen such that no two lie on a line parallel to any side of the large triangle must be less than 8.
   - By similar reasoning, it can be shown that choosing 7 vertices does not lead to a contradiction.

The final answer is \(\boxed{7}\).