Let $O$ and $L$ be the circumcenter and symmedian point. We reconize A_1 as the $A$-dumpty point. Hence $LA_1O=90^{\circ}$ so $A_1\in(OL)$. Similarly, $B_1\in(OL)$ and $C_1\in(OL)$ so $O_1$ is the midpoint of $OL$. Now from USATSTST 2015 P2(https://artofproblemsolving.com/community/c6h1106812p5017915) we know $O_2$ is the midpoint of $HG$ where $G$ is the centroid and $H$ is the orthocenter. Hence $MO_1=0.5\cdot GO=0.5\cdot GO_2=NO_2$, and $MO_2=0.5\cdot HL=NO_1$ so $O_1MO_2N$ is a parallelogram.

Attachments:

Similar to above. Let $O$ be the circumcenter of $\triangle ABC$, $K$ the symmedian point, $G$ the centroid, and $H$ the orthocenter. Note that as $\angle OA_1K=\angle OB_1K=\angle OC_1K=90^\circ$, $A_1,B_1,C_1$ lie on the circle with diameter $OK$. In particular, $O_1=\frac{O+K}2$. We show that $A_2$ lies on the circle with center $HG$. To do so, let $P$ be the midpoint of $BC$, and we claim that $PA_0$ is the reflection of $AP$ over $BC$. Indeed, note $(AA_0;BC)=-1$, so if $T=BB\cap CC$, we know that $(AA_0;T(AA_0\cap BC))=-1$, so thus we know that $BC$ is the angle bisector of $\angle APA_0$. In particular, $$\measuredangle AA_0P=\measuredangle A_0AP+\measuredangle APA_0=2\measuredangle PAB-\measuredangle CAB+2\measuredangle APB=\measuredangle CBA-\measuredangle ACB$$Let $H’$ be the reflection of $H$ over $BC$. Then, $\measuredangle AA_0H’=\measuredangle ACH’=90^\circ+\measuredangle ACB-\measuredangle CBA$, so $\angle H’A_0M=90^\circ$. Reflecting over $BC$ and noting $BC$ is the angle of $\angle APA_0$, hence $PA_0$ maps to the $A$-median. Thus, $\angle HA_2G=90^\circ$, as desired. The conclusion follows as $O_1=\frac{O+K}2,M=\frac{G+K}2,N=\frac{3G-O}2,O_2=\frac{H+G}2$, so $$O_1+O_2=\frac{O+K+H+G}2=\frac{4G-O+K}2=M+N$$