4. Given are convex quadrilaterals $A B C D$ and $P Q R S$, where the vertices of quadrilateral $P Q R S$ lie on the sides or inside the quadrilateral $A B C D$. Can the sum of the diagonals of quadrilateral $P Q R S$ be greater than the sum of the diagonals of quadrilateral $A B C D$?
4. The answer is confirmed. Let the quadrilateral $ABCD$ be a rhombus with vertices $A(1,0), B(0,50), C(-1,0)$, and $D(0,-50)$. Let points $P$ and $Q$ be the intersections of the line $y=49$ with the rhombus $ABCD$ (where $P$ is to the right of $Q$), and points $R$ and $S$ be the intersections of the line $y=-49$ with the rhombus $ABCD$ (where $R$ is to the left of $S$). Then $PQRS$ is a rectangle, and for the sum of its diagonals, we have $PR + QS > 2PS = 2 \cdot 98 = 196$, but $AC + BD = 2 + 100 = 102 < 196$.