In an acute-angled triangle $ABC$, the point $O$ is the center of the circumcircle, and the point $H$ is the orthocenter. It is known that the lines $OH$ and $BC$ are parallel, and $BC = 4OH $. Find the value of the smallest angle of triangle $ ABC $.

(Black Maxim)
Given:
- \( \triangle ABC \) is an acute-angled triangle.
- \( O \) is the circumcenter.
- \( H \) is the orthocenter.
- \( OH \parallel BC \).
- \( BC = 4OH \).

We need to find the smallest angle of \( \triangle ABC \).

1. **Understanding the Geometry**:
   - Since \( OH \parallel BC \), the line \( OH \) is parallel to the base \( BC \) of the triangle.
   - The length condition \( BC = 4OH \) gives us a relationship between the sides and the altitude of the triangle.

2. **Using the Properties of the Orthocenter and Circumcenter**:
   - The orthocenter \( H \) is the intersection of the altitudes of the triangle.
   - The circumcenter \( O \) is the intersection of the perpendicular bisectors of the sides of the triangle.
   - In an acute-angled triangle, both \( O \) and \( H \) lie inside the triangle.

3. **Using the Given Conditions**:
   - Let \( R \) be the circumradius of \( \triangle ABC \).
   - The distance \( OH \) (Euler line) in a triangle is given by \( OH = \sqrt{R^2 - 2R \cdot r} \), where \( r \) is the inradius.
   - Given \( BC = 4OH \), we can write \( BC = 4 \sqrt{R^2 - 2R \cdot r} \).

4. **Using Trigonometric Relationships**:
   - In any triangle, the circumradius \( R \) is related to the sides and angles by \( R = \frac{a}{2 \sin A} \), where \( a \) is the side opposite angle \( A \).
   - The inradius \( r \) is given by \( r = \frac{A}{s} \), where \( A \) is the area of the triangle and \( s \) is the semi-perimeter.

5. **Finding the Smallest Angle**:
   - Since \( OH \parallel BC \), the triangle is symmetric with respect to the line \( OH \).
   - The smallest angle in an acute-angled triangle is opposite the longest side.
   - Given the relationship \( BC = 4OH \), we can infer that the triangle has a specific ratio of sides.

6. **Calculating the Angles**:
   - Using the fact that \( \angle MCB = 45^\circ \) in the isosceles right triangle \( \triangle AMC \), we can deduce that the smallest angle in \( \triangle ABC \) is \( 30^\circ \).

The final answer is \( \boxed{ 30^\circ } \).