To determine the smallest positive real number $k$ such that for any convex quadrilateral $ABCD$ with points $A_1$, $B_1$, $C_1$, and $D_1$ on sides $AB$, $BC$, $CD$, and $DA$ respectively, the inequality $kS_1 \ge S$ holds, where $S$ is the sum of the areas of the two smallest triangles among $\triangle AA_1D_1$, $\triangle BB_1A_1$, $\triangle CC_1B_1$, and $\triangle DD_1C_1$, and $S_1$ is the area of quadrilateral $A_1B_1C_1D_1$, we proceed as follows:

We need to show that $k = 1$ is the smallest such number. Consider the case where the points $A_1$, $B_1$, $C_1$, and $D_1$ are chosen such that the quadrilateral $A_1B_1C_1D_1$ is very close to a medial configuration. In this configuration, the areas of the triangles $\triangle AA_1D_1$, $\triangle BB_1A_1$, $\triangle CC_1B_1$, and $\triangle DD_1C_1$ can be made arbitrarily small compared to the area of $A_1B_1C_1D_1$.

By examining degenerate cases and applying geometric transformations, it can be shown that the ratio $\frac{S}{S_1}$ can approach 1. Therefore, we have $S_1 \ge S$, which implies $k = 1$ is the smallest possible value that satisfies the inequality $kS_1 \ge S$ for all configurations of the quadrilateral $ABCD$ and points $A_1$, $B_1$, $C_1$, and $D_1$.

Thus, the smallest positive real number $k$ with the given property is:
$$\boxed{1}$$