Two helium balloons meet a cocktail party. Do you think they might find each other attractive? Here we are talking about the mutual force between the two balloons. This is in the context of Newton's law of gravity (so electrostatics etc. are ignored). You can assume that the cocktail party is held in a room that contains air. It is not necessary to assume that the Earth's gravitational field is present, but if it were present, the helium balloons would tend to rise to the ceiling.
[Edit] Question was re-posted:
Is the force due to gravity between two objects immersed in a fluid correctly given by$ -G(M_1-m_1)(M_2-m_2)/r^2$? ($m$ = mass of displaced fluid)
 A: The force due to gravitational effects between two spherical objects, A and B, immersed in a fluid is  -G(MA-mA)(MB-mB)/r2 where M is the mass of an object and m is the mass of the fluid it displaces. 
For two helium balloons, both (MA-mA) and (MB-mB) will be negative, so the balloons will find each other attractive.  If one of the balloons is filled with CO2, they will find each other quite repulsive.  If both balloons are filled with CO2, they will again be attracted to each other. 
[It's probably bad form to answer one's own question, but I wasn't aware of this result and I thought it was rather cute] 
I'm editing this to add a proof since there has been some scepticism expressed about the formula: 
Proof:

*

*The attraction between the two balloons is measured as the change in total force on the first balloon, A, when the second balloon, B, is introduced.

*The introduction of the second balloon changes the mass distribution and the gravitational fields and the pressure gradients in the air.

*The change in mass distribution is such that a spherical volume of helium with mass, MB, replaces a spherical volume of air with mass, mB. The change in mass is MB - mB.

*As a result of the change is mass, the gravitational field in the surrounding air also changes. The change in field is $\Delta$g = -G(MB-mB)/r2, where r is the distance from the center of the sphere.

*Note: the change in field also causes a change in the pressure gradients in the air.

*As a result, there is a change in force on the first balloon, A.  The force has two components, one arising from the change in gravitational force on balloon A and the other arising from the change in pressure gradient across the balloon A's surface (i.e. it's buoyancy).

*According to Newton, the first force is $\Delta$g.MA.  According to Archimedes, the second force is -$\Delta$g.mA where, again, mA is the mass of the displaced air. The net change in force is $\Delta$g(MA-mA).

*Substituting $\Delta$g, gives us the final version of the formula. The force between the two (spherical) helium balloons in air is -G(MA-mA)(MB-mB)/r2 operating along the line joining their centers.

