smc

Construct triangle $△ABC$ with points $M,N,P$ being the midpoints of sides $\overline{CB},\overline{AB},\overline{AC}$, respectively. Proceed by drawing all medians. Then draw all medians (so draw $\overline{AM},\overline{BP},\overline{CN}$). Next, draw line $\overline{PM}$ and label $\overline{PM}$'s intersection with $\overline{CN}$ as the point $Q$. From the problem, the area of $△QMO$ is $n$, but by vertical angles we know that $\angle QOM=\angle AOM$. Furthermore, since line $\overline{PM}$ is drawn from the midpoint of $\overline{AC}$ to the midpoint of $\overline{CB}$, we know that $\overline{PM}$ is parallel to $\overline{AB}$ (via SAS similarity on triangles PCM and ABC). From these parallel lines we know that $\angle PMO=\angle OAB$ which indicates that $△QOM\sim △ANO$. The linear ratio from $△QOM$ to $△ANO$ is 1:2 because line segment $\overline{OM}$ is one half of line segment $\overline{AO}$ since $\overline{AO}$ and $\overline{OM}$ make up the median $\overline{AM}$. Thus the area ratio is 1:4. So $△ANO$ has area $4n$. Since $△ONB$ has the same height and base as $△ANO$ we know that the area of $△AOB=8n$. The medians form 3 triangles each with area $1/3$ of the total triangle (these triangles are $△AOB,△COB,△COA$). Thus since $△AOB=4n\underset{\text{multiply by 3}}{⟹}△ABC=24n$ $\boxed{24n}$.