It seems that I created a mess by switching ##J_1## and ##J_2## in post #30. It behooves me to offer my apologies and then clean up my mess. Here are my revised arguments.
Assume a model in which the "stick" lies on its side on a frictionless surface and that impulses ##J_1=mg\Delta t## and ##J_2=F\Delta t## are delivered simultaneously as shown in the diagram, post #30. Consider the y-component of the velocity of point P. As
@haruspex pointed out in post #51, there is no normal force at that point in the model but there is in the actual problem. Below I explain why I think the model is applicable.
The normal force is what is necessary to provide the observed acceleration. Since point P in the model starts from rest, the observed acceleration will be in the direction of the initial velocity. If in the model the initial velocity is in the negative y-direction, in the actual problem the normal force will increase above the static value of ##\frac{1}{2}mg.## Of more interest is the condition(s) under which the normal force will decrease and may even become to zero. This possibility could be realized if in the model the initial velocity of point P is in the positive y-direction. Then in the actual problem the normal force could be diminished below the static value of ##\frac{1}{2}mg##, perhaps all the way down to zero.
The condition for positive vertical velocity at P is $${J_2}(1-\beta)-{J_1} >0~~~~\left(\beta \equiv \frac{3x^2}{x^2+y^2}\right).$$Let us consider a rod of fixed length ##2x## and variable height ##2y##. The range of values for ##\beta## is ##0<\beta<3.## We substitute ##J_1=mg\Delta t## and ##J_2=F\Delta t## to get $$F(1-\beta) >mg~~~~(0<\beta<3).\tag{1}$$Inequality (1) is a
condition that must be satisfied for the normal force to drop below the static value. It can be satisfied only in the range ##0<\beta<1## because ##mg## cannot be less than a negative number.
Specifically, note that ##\beta =0## corresponds to ##y\rightarrow \infty## at constant ##x## which is essentially describing a vertical stick standing on its tip. Then ##F=mg## is the threshold for the condition to be satisfied. Well, we know that the pulling force must be greater than the weight of the stick in order to lift it off the surface (##N=0##) so there is nothing new here. The CM is already above the "corner" P, there is no rotation of P about the CM, therefore P can only go up when the ##F## is greater than the weight.
What about sticks such that ##0<\beta<1##? From post #24 we have the equation $$N(\beta+1)=mg+F(\beta-1)\implies N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right].\tag{2}$$We want to know what ##N## looks like when condition (1) is satisfied. If we replace ##F(1-\beta)## in the RHS of equation (2) with the smaller term ##mg## from inequality (1), then LHS > RHS: $$N>\frac{1}{1+\beta}(mg-mg)\implies N>0.$$It looks like the normal force will be positive and hence non-zero for all values of ##\beta##. All this agrees with one's intuition. As the stick is lifted off the surface, point P rotates about the CM and pushes
down on the surface thus maintaining a non-zero normal force. The normal force will not go to zero unless the CM is above P in which case the stick is lifted straight off the surface (##N=0##). The actual problem asks for the normal force "shortly after the right end of the stick leaves the surface". This happens before the CM and point P are lined up along the vertical unless, as we have seen, the stick is thin and already along the vertical.