For refrence: In the figure shown:
H ➔ Orthocenter of $\triangle$ ABC; also: BH = 16, AC = 30 and BN = NC
Find: MN
An important property that can be useful: $OG = 2BH \therefore BH = 8$ Tracing the circumference circumscribed to the triangle $QBC$ we can get the value of $BC$ and therefore the value of $BN$ and $NC$ and also of $CQ$.
$M$, the midpoint of $OH$, is the centre of the nine-point circle. This circle passes through the midpoints of the sides of the triangle. Therefore, $MN=$ radius of the nine-point circle$=$ $\frac {1}{2}\cdot $ $($ Circumradius of $\triangle ABC$ $)=\boxed {\frac {17}{2}}$ $($Apply Pythagorean Theorem on $\triangle OGC$ $)$.