Is it possible?

From symmetry we can say that both the beads at any instant will make equal angle the vertical.

Draw the FBD of any bead when it makes angle $\theta$ with the vertical. In the diagram assume that ring provides normal force ( $N$ ) towards the center to the bead(and not away from the center).

Taking radial acceleration of the bead to be $0$ we can write that $mg\cos\theta + N = \dfrac{mv^{2} }{R}$ By conservation of energy we can write that:- $\dfrac{1}{2}mv^{2}= mgR(1-\cos\theta)$ $v= \sqrt{2gR(1-\cos\theta)}$ From the previous eqaution we can write that:- $N + mg\cos\theta = \dfrac{m.2gR(1-\cos\theta)}{R}$ $N =2mg-3mg\cos\theta$ This is the same force which the bead is providing to the ring in the direction making angle $\theta$ with the ring.

Let total force provided by the beads is $F_{net}$ . $F_{net} = 2N\cos\theta$ $F_{net} = 2(2mg - 3mg\cos\theta)\cos\theta$ $F_{net} = 2mg(2-3\cos\theta)\cos\theta$ For Maximum $F_{net}$ :- $((2-3\cos\theta)\cos\theta)' = 0$ $\cos\theta = \dfrac{1}{3}$ Hence:- $F_{net} =\dfrac{2mg}{3}$ This force should be equal to $Mg$ to lift the ring:- $\dfrac{2mg}{3} = Mg$ $\dfrac{m}{M} = \dfrac{3}{2}$