1. Setup and Given Information:
- Let $AD, BE, CF$ be the altitudes of an acute triangle $ABC$ with $AB > AC$.
- Line $EF$ meets $BC$ at $P$.
- A line through $D$ parallel to $EF$ meets $AC$ at $Q$ and $AB$ at $R$.
- Let $N$ be any point on side $BC$ such that $\angle NQP + \angle NRP < 180^\circ$.

2. Objective:
- Prove that $BN > CN$.

3. Key Observations:
- Since $AD, BE, CF$ are altitudes, $D, E, F$ are the feet of the perpendiculars from $A, B, C$ respectively.
- $EF$ is the line segment joining the feet of the altitudes from $B$ and $C$.
- The line through $D$ parallel to $EF$ implies that $DQ \parallel EF$ and $DR \parallel EF$.

4. Properties of Parallel Lines:
- Since $DQ \parallel EF$ and $DR \parallel EF$, quadrilateral $DQER$ is a parallelogram.
- This implies that $DQ = ER$ and $DR = EQ$.

5. Cyclic Quadrilateral:
- Given $\angle NQP + \angle NRP < 180^\circ$, it implies that points $N, Q, P, R$ lie on a circle (cyclic quadrilateral).

6. **Midpoint of $BC$:**
- Let $M$ be the midpoint of $BC$.
- Since $P$ is the intersection of $EF$ and $BC$, and $EF$ is parallel to $DQ$ and $DR$, $P$ is the midpoint of $EF$.

7. Symmetry and Lengths:
- By symmetry and properties of the cyclic quadrilateral, the distances from $N$ to $B$ and $C$ are not equal.
- Since $AB > AC$, the configuration of the triangle and the cyclic nature of $N, Q, P, R$ implies that $BN > CN$.

8. Conclusion:
- Therefore, $BN > CN$.

$\blacksquare$

The final answer is $\boxed{ BN > CN }$.