Finite probability from discrete distribution and zero probability from continuous distribution in Maxwell-Boltzmann statistics
The probability of any one specific velocity is not zero, it's just very, very small.  Infinitismal, or tending towards zero, is how we usually put it.  The reason is because the spacing between allowed translational states is inversely proportional to the size of the container, and when you are talking about a macroscopic container (a bottle of gas, say, or beaker of liquid) then from the atomic point of view this is gargantuan and the spacing between allowed translational energies very, very small.  (If you want to get a feel for this, look up the allowed energies for a particle in a box.  You'll see that the size of the box goes in the  denominator and as the box gets bigger the spacing between energy levels gets smaller and smaller.  You can amuse yourself by calculating the exact difference between the 1st and 2nd energy levels for, say, an H atom in a box 1cm x 1 cm x 1cm.  It's really small!)

So small that to do the sum over states in your top formula is a gigantic pain, quite challenging.  But if you convert it to an integral, the integral can be done pretty easily.  So that's what we do.  We replace a huge sum, over 10^20 terms or more, each of size 10^-20 or less, with an integral.  Think of it as the reverse of when you learned integral calculus, and in order to figure out an integral you replaced it with a large sum -- and let the size of each term go to zero while the number of terms went to infinity.

Here, you are taking a sum which has a very large number of terms, each of which is very small, and replacing it with an integral.  It's not exact, but it's very good, and it becomes perfect in the limit of an infinite-sized container, and anyway the errors for ordinary-sized containers are ridiculously small, unmeasureable.