If molecules in the sky scattered red light instead of blue light, the sunsets would be blue.
A Longer Answer:
First, it's helpful to understand why the sky is blue and sunsets are red.
Second, it's helpful to be accurate: It's the atmosphere, not the "sky" (where the stars are) which is blue during the day and which is red at sunset. The sky is black--ask any space traveler!
This kind of scattering is caused mainly by photons interacting with the nitrogen and oxygen molecules in the earth's atmosphere. Lord Rayleigh determined the formula that describes the scattering of light by air molecules that are smaller than the wavelength of light.
The formula says that the magnitude of scattering is inversely proportional to the 4th power of the wavelength λ. That means that blue light, with λblue ~= 400nm ends up being scattered about 10X more than red light, with λred ~= 700nm.
(The white light from the sun ranges across the visible light spectrum, with blue and red on each end.)
What is interesting--and often confusing at the beginning--is that this λ^-4 Rayleigh scattering explains BOTH why the sky is blue at midday and why sunrises and sunsets are red!
How this happens is explained at some level at Reference 1. Basically, it has to do with how many nitrogen and oxygen molecules the sunlight must pass through before it reaches our eyes. At noon, the thickness of the atmosphere between our eyes and the sun is at a minimum; as the sun approaches sunrise and sunset, that thickness changes very rapidly.
What actually happens in air is that the photons at the blue end of the spectrum are much more susceptible to scattering (by the O2 and N2 molecules) than are the red photons--Rayleigh's equation shows this. When looking upwards into the atmosphere during daylight, the atmosphere is just the right depth so that, even though the blue photons are scattered, enough of them still reach your eye from every direction to make the atmospheric molecules look blue! The rest of the sunlight that is NOT scattered comes to your eye on a direct line from the sun (be careful!) as (mostly) white light, or as reflected from the surfaces of the things we see.
Now, what happens at sunset? Well, the light from the sun must get past many, many more oxygen and nitrogen molecules. Again, ONLY the blue photons are scattered--the remaining photons are unaffected by the λ^-4 Rayleigh and proceed directly to your eye. (There are other types of scattering and absorption that become more prominent as the atmosphere "gets thicker"--we'll ignore those here.) The PROBLEM is that there are so many more scattering molecules, that few if any of the blue photons reach your eye! In fact, almost none of the photons from the blue half of the spectrum reach your eye, because they just keep being scattered and scattered.
So, if none of the blue-end photons reach your eyes at sunset, how do you still see the sun? And the answer, of course, is because the reddish photons are not scattered and come directly to your eye, so the sun looks red. The nearby clouds look red because they are reflecting the red light from the sun.
That's the background for the answer to the ORIGINAL question: If molecules in the atmosphere scattered red light instead of blue light, what color would the sunsets be?
This means that we're inverting the Rayleigh scattering rule from λ^-4 (lambda-to-the-MINUS-4-power) to λ^+4 (lambda-to-the-PLUS-4-power). This in turn means that the higher-wavelength red end of the spectrum is scattered more than the blue end. So, in this inverted world, it's the BLUE photons which make it through the N-O molecular scattering maze instead of the red, which just keep getting scattered over and over until they lose all their energy and never reach your eyes.
So, the answer is the same as the Short one: The sunsets would be blue, not red.
(Another useful reference, which gives the Rayleigh formula, is number 2.)