$(YUG 3)$ Let four points $A_i (i = 1, 2, 3, 4)$ in the plane determine four triangles. In each of these triangles we choose the smallest angle. The sum of these angles is denoted by $S.$ What is the exact placement of the points $A_i$ if $S = 180^{\circ}$?
1. **Initial Setup and Assumptions:**
   Let the four points be \(A_1, A_2, A_3, A_4\) and denote them as \(A, B, C, D\) for convenience. We need to determine the placement of these points such that the sum of the smallest angles in the four triangles formed by these points is \(S = 180^\circ\).

2. **Non-Convex Quadrilateral Case:**
   If the four points do not form a convex quadrilateral, assume without loss of generality that point \(A\) is inside the triangle \(BCD\). In this case, the sum of the smallest angles in the triangles \(ABC, ABD, ACD, BCD\) is less than \(180^\circ\):
   \[
   S \leq \angle ABC + \angle ABD + \angle ADC + \angle BCD < 180^\circ
   \]
   Therefore, \(ABCD\) must be a convex quadrilateral.

3. **Convex Quadrilateral Case:**
   Assume \(ABCD\) is a convex quadrilateral. Let the diagonals \(AC\) and \(BD\) intersect at point \(E\). Without loss of generality, assume \(\angle AEB \geq 90^\circ\). Then:
   \[
   S \leq \angle CAB + \angle DBA + \angle ACD + \angle BDC = 2\angle BEC \leq 180^\circ
   \]
   Equality holds if \(AC \perp BD\).

4. **Further Analysis:**
   Assume \(CE \geq AE\) and \(BE \geq DE\). Then:
   \[
   \angle EDC \geq 45^\circ \geq \angle ABD
   \]
   Therefore:
   \[
   S \leq \angle ACB + \angle ACD + \angle ABD + \angle CBD \leq 180^\circ
   \]
   For equality to hold, \(AB \parallel CD\).

5. **Conclusion:**
   Since \(DE \leq BE\), we have \(CE \leq EA\), so \(CE = EA\) and \(BE = DE\). Hence, \(ABCD\) is a rhombus. It is easy to check that all rhombuses work, except when one of the angles is less than \(60^\circ\).

The final answer is \( \boxed{ ABCD } \) must be a rhombus with all angles at least \(60^\circ\).