Best regards,
Bui Quang Tuan

Suppose (O') is circumcirle of triangle QM4N4.

Y = intersection of line M1M2 (Lm) with MN (L1)
Z = intersection of line N1N2 (Ln) with MN (L1)
Denote circumcircles of MXQ and NXQ as (Oy), (Oz) respectively.

Y is intersection of M1M2, MX therefore Y is radical center of (O), (Om) and (Oy). It means Y is on common chord of (O), (Om). (1)
Y is intersection of QM4, MX therefore Y is radical center of (O'), (Om) and (Oy). It means Y is on common chord of (O'), (Om). (2)
Radical center of (O), (O') and (Om) is intersection of common chord of (O), (Om) and common chord of (O'), (Om). By (1) and (2) it is point Y. So Y is radical center of (O), (O') and (Om). It means: common chord of (O), (O') passes through Y.
Similarly, we can show that common chord of (O), (O') passes through Z.
Because Y, Z is on L1 so L1 is common chord of (O), (O'). Therefore M3, N3 are intersections of two circles (O), (O'). It means Q, M3, N3, M4, N4 are concyclic on (O')

Now it is enough to combine Six Concyclic Points with Five Concyclic Points theorems in one configuration with Twelve Concyclic Points.

The idea is very natural. Suppose P is Cevian point of triangle ABC and three circles (Oa), (Ob), (Oc) passing P with other common points A'B'C' on Cevians PA, PB, PC. By Six Concyclic theorem: if these three circles intersect ABC sidelines then intersections are concyclic. We can see that P is also Cevian point of triangle A'B'C' and three circumcircles PAB, PBC, PCA also can intersect sidelines of A'B'C at concyclic points. So we have two six concyclic points: one six points on sidelines of ABC, other six points on sidelines of A'B'C'. If we choose proper configuration of ABC, A'B'C', P such that these two concylic circles are same one then we have 12 concyclic points.

I suggest here one case and it is very famous configuration: ABC and A'B'C' are two orthologic triangles. Detail as following:

ABC and A'B'C' are two orthologic triangles. It means that three perpendicular lines from A, B, C to B'C', C'A', A'B' respectively are concurrent at one point, say P. Suppose Ap, Bp, Cp are intersections of PA, PB, PC with B'C', C'A', A'B' respectively. We construct three circles (Oa), (Ob), (Oc) as circumcircles of PA'BpCp, PB'CpAp, PC'ApBp. By Six Concyclic theorem these three circles intersect sidelines of ABC at six concyclic points. Denote concyclic circle as (O). Please see attached file!
It is well known that three perpendicular from B', C', A' to BC, CA, AB respectively are also concurrent at one point, say Q. Suppose Aq, Bq, Cq are intersections of QA', QB', QC' with BC, CA, AB respectively. By the same way, we construct three circles (O'a), (O'b), (O'c) as circumcircles of QABqCq, QBCqAq, QCAqBq. By Six Concyclic theorem these three circles intersect sidelines of A'B'C' at six concyclic points. Denote concyclic circle as (O').
Now we use Five Concyclic Points theorem to prove that (O) and (O') are the same one.
The Five Concyclic Points configuration can be see at each vertex A, B, C, A', B', C'. For example at A.
B'C' is line L1, AP is line L2, Ap is point X, A is point Q, AC is Lm, AB is Ln, (Ob) is (Om), (Oc) is (On),... So by Five Concyclic Points theorem: two intersections of (O) with B'C' are on circle (O'a), circumcircle of QABqCq.
Similarly at other B, C vertices, we can show that all concyclic points of (O') are on (O). At the end, we have total 12 concyclic point. It is easy to show that center of concylic circle is midpoint of PQ.

Best regards,
Bui Quang Tuan