For convenience, we let $A_1,...,A_4$ be $A,B,C,D$. If the 4 points does not form a convex quadrilateral, then WLOG let $A$ be inside $BCD$. Then $S\leq \angle ABC+\angle ABD+\angle ADC+\angle BCD<180$, so $ABCD$ must be a convex quadrilateral. Let the diagonals $AC,BD$ intersect at $E$. WLOG let $\angle AEB\geq 90$. Then $S\leq \angle CAB+\angle DBA+\angle ACD+\angle BDC=2\angle BEC\leq 180$ with equality if $AC\perp BD$. Now WLOG let $CE\geq AE$ and $BE\geq DE$. Then $\angle EDC\geq 45\geq ABD$, so $S\leq \angle ACB+\angle ACD+\angle ABD+\angle CBD\leq 180$ so equality must hold so $AB\parallel CD$. But since $DE\leq BE$, we have $CE\leq EA$ so $CE=EA$ so $BE=DE$. Hence $ABCD$ is now a rhombus, and it is easy to check that all rhombus works, except when one of the angle is less than 60.