I linked a question which does also answer this question, however, on second thought, I think an answer with a different approach can be helpful.
If we suppose that only two particles are produced, then as you say, the energy released in the decay must be shared between those two particles. This is energy conservation. We can also consider the original nucleus to be at rest, that is, with zero momentum. Then by momentum conservation, the total momentum of the two particles must also be zero. You can easily write down the algebraic formulae that represent this. They are also found in the linked question.
But let's just think about the possibilities without doing algebra. Each particle has some momentum which is a 3-component vector, for 3-dimensional space. The energy of each particle is fixed once we know its momentum and mass. So then, if we have two particles, we need $2\times 3=6$ numbers to describe their kinematics. But, we already said that the particles have a fixed total energy. That means once I know 5 of those numbers, I could calculate the 6th by requiring the total energy to be fixed. So now I only have 5 numbers that I can choose freely. And I also have to be sure that the total momentum is zero. That's three requirements, because it must be true for each direction. Imposing those constraints, I'm now down to $6-1-3=2$ numbers that I could choose. But actually, if you think about it, it doesn't really change anything if I rotate the whole situation in space. It's the same process in any case. So let me take the direction of the $\beta$-particle, and pick a certain direction for it to point. That's a choice of two numbers, a latitude and a longitude, if you like. So now I've fixed $1+3+2=6$ conditions on $6$ numbers describing the momentum of my two particles, so actually there's absolutely no freedom left whatsoever. Everything is already determined. So, even without working through the algebra, you can see that with only two particles, the energies and momenta must come out the same every time. There's no freedom for you to share the energy differently between them.
But, if you add a third particle, now you have $3\times 3=9$ numbers to describe the momenta. Again you have 1 energy and 3 momentum conservation conditions. And now you actually have a bit more choice with the overall rotation. You can pick a latitude and longitude for the $\beta$, and then a rotation around that direction for, say, the neutrino. So that's three rotations that aren't important because they don't really change the situation. Now if we do our accounting, we get $9-1-3-3=2$. So there are still two unknowns, two numbers that we can pick freely and get a truly different outcome. So now the energy of the $\beta$ can be in some range, depending the values of those two unfixed numbers. You can do the algebra to find out exactly what that range is, but we already see why there is a major qualitative difference between the two particle and three particle cases. And as soon as you see any range of energies for the $\beta$, you know (if you believe energy/momentum conservation, as Pauli was willing to do) that there must be at least 3 particles, even without doing any calculations.
By the way, dear radioactive ladies and gentlemen, the letter in which Pauli proposes the existence of the neutrino is a must-read. Also, if Pauli can say he's skipping some of his professional responsibilities because there's a party that can't go on without him, so can you.
