Solution:

The answer is no. Assume that, on the contrary, it is possible to partition a polygon $P$ into non-convex quadrilaterals. Let $n$ be the number of quadrilaterals. Denote by $S$ the total sum of all internal angles of all the quadrilaterals. Since the sum of internal angles of each quadrilateral is $360^{\circ}$, we have $S = 360^{\circ} n$. However, each of the non-convex angles has to be in the interior of $P$, hence the sum of angles around the vertex of that angle has to be $360^{\circ}$. This immediately gives $360^{\circ} n$ as the sum of angles around such vertices. Since those are not the only vertices (at least the vertices of $P$ will contribute to the sum $S$), we have that $S > 360^{\circ} n$ and this is a contradiction.