Teasing Quadrilateral

First, note that side lengths of $2,3,3,4$ give rise to a convex quadrilateral (a trapezoid) that is not a rhombus or a parallelogram, and doesn't have 3 equal sides. So the only possible answers are "at least 2 sides are equal" or "all sides can be distinct." We will show that in fact, at least 2 sides must be equal.

If the sides are $x,y,z,w,$ then the condition is that each of the sides divides the perimeter $x+y+z+w.$ Suppose there is such a quadrilateral with all distinct sides. Suppose without loss of generality that $x<y<z<w.$ Now if $x+y+z+w=2w,$ then the quadrilateral is degenerate. And $x+y+z+w < w+w+w+w = 4w.$ So the sum must equal precisely $3w.$

Let $n=3w$ be the perimeter. Then $x = n/a, y = n/b, z = n/c, w = n/3,$ for distinct positive integers $a,b,c > 3.$ So we get $n = \frac{n}3 + \frac{n}{a} + \frac{n}{b} + \frac{n}{c},$ and the largest that the right side can possibly be is $\frac{n}3 + \frac{n}4 + \frac{n}5 + \frac{n}6 = \frac{19n}{20},$ so this is impossible.

Hence at least 2 sides are equal.