I'm not at my office to verify this, but I think the software will divide the 500 pounds over the area regardless of how many surfaces are selected.  My understanding of the application of a surface load is that it will calculate the pounds/in^2 based on the surface area selected.

The question may be in how the physical load is being applied in the real world.  For instance, if you have 5 different surfaces each being pressed upon with a 100 pound load and the surfaces are different sizes, then you need to apply the 5-100 pounds loads.  If, however, you have 5 surfaces reacting a 500 pound load, then you should be able to select all 5.

It comes down to the total load you need applied to a particular surface.  If you need larger surfaces to take more of the load, a surface pressure will do this when all 5 surfaces are selected.  5 surfaces totaling 100 in^2 would result in 5psi, but that means that a surface having 20 in^2 would carry twice the load (100 pounds) of a surface with only 10 in^2 (50 pounds).  On the other hand, applying 100 pounds to each of these two surfaces would result in the smaller surface carrying a higher psi loading, but an equivalent force (the 20 in^2 surface would carry 100 pounds as 5psi while the 10 in^2 surface would carry the load as 10psi).

A small, simple model should verify this.

Garland E. Borowski, PE
Borowski Engineering & Analytical Services, Inc.
Lower Alabama SolidWorks Users Group
Magnitude The Finite Element Analysis Magazine for the Engineering Community