Let $ABC$ be an acute-angled triangle with altitude $AT = h$. The line passing through its circumcenter $O$ and incenter $I$ meets the sides $AB$ and $AC$ at points $F$ and $N$, respectively. It is known that $BFNC$ is a cyclic quadrilateral. Find the sum of the distances from the orthocenter of $ABC$ to its vertices.
1. Let $ABC$ be an acute-angled triangle with altitude $AT = h$. The line passing through its circumcenter $O$ and incenter $I$ meets the sides $AB$ and $AC$ at points $F$ and $N$, respectively. It is given that $BFNC$ is a cyclic quadrilateral.
2. We need to find the sum of the distances from the orthocenter $H$ of $ABC$ to its vertices.
3. Let $AI \cap BC = X$ and $D$ be the foot of the perpendicular from $I$ to $BC$. Let $r = ID$ be the inradius and $R = AO$ be the circumradius.
4. Without loss of generality, assume $\angle B \ge \angle C$. It is given that $FN$ is antiparallel to $BC$ with respect to $\angle BAC$. Also, $AO$ is the isogonal conjugate of $AT$ with respect to $\angle BAC$. Hence, $AO \perp IO$.
5. Note that:
   \[
   \angle IAO = 90^\circ - \angle C - \frac{\angle A}{2} = \frac{\angle B - \angle C}{2}
   \]
   \[
   \angle DIX = \angle TAX = \frac{\angle A}{2} - (90^\circ - \angle B) = \frac{\angle B - \angle C}{2} = \angle IAO
   \]
6. Therefore, $\triangle AIO \sim \triangle IDX \sim \triangle ATX$. So, we have:
   \[
   \frac{r}{h} = \frac{IX}{AX} \quad \text{and} \quad \frac{R}{h} = \frac{AI}{AX}
   \]
7. Adding these ratios, we get:
   \[
   \frac{R + r}{h} = \frac{AI + IX}{AX} = 1 \implies R + r = h
   \]
8. Denote the orthocenter by $H$. The required sum of the distances from the orthocenter to the vertices is:
   \[
   AH + BH + CH = 2(R + r) = 2h
   \]

The final answer is $\boxed{2h}$.