Re: Why is the Kf value for a solution always greater than the Kb value?

A small amount of involatile dissolved matter will (nearly?) always make a 
much larger difference to the freezing point of a liquid than to its boiling 
point.

Is seems unlikely that there are any exceptions, but I cannot be completely 
sure. If there are, they are likely to be solvents like liquid hydrogen or 
liquid neon.
Why is this the case?

You have gone into the physical chemistry textbooks enough to know about Kf 
and Kb. If you go one step further, and check in, for example, Atkins 
"Physical Chemistry" (around page 169 in my Fourth Edition), you will find 
out that 

Kb = R * Tb^2 / delta H evap      and 
Kf = R * Tf^2 / delta H fus

where Tb and Tf are the boiling and freezing temperatures, and delta H evap 
and delta H fus are the molar latent heats of evaporation and fusion. Heats 
of evaporation are normally much greater than heats of fusion. But that is 
just pushing your question back one level. Why?

It is useful to go yet one step further again to find that the molar 
entropies of fusion and evaporation are equal to the corresponding latent 
heats divided by the boiling and freezing temperatures respectively.

SO

Kb = R * Tb / delta S evap     and
Kf = R * Tf / delta S fus.

A quick glance at standard entropy tables will tell you that stamdard 
entropies of gases are always much larger than those of solids or liquids. 
If you think of entropy as representing "disorder" in some sense, you will 
understand that the main factor is the huge increase in molar volume in 
going from a condensed phase to a gas at 1 atmosphere. Most solid/liquid 
transitions involve a volume increase of only 10-20% at most. This is the 
main factor that normally makes entropy of evaporation hugely greater than 
entropy of fusion -- more than enough to outweigh the difference between 
melting and boiling temperatures.