Understanding why there is such a voltage is simple, if one takes care to what we are speaking of.
An answer
An answer could be : Power loss is \$I^2\times R_{wire}\$. So for a constant transmitted power \$V\times I\$, by increasing voltage we decrease \$I\$, thereby diminishing loss.
Ok, but did we not just hide our \$V\$ variable ?
Indeed, by Ohm's Law \$V = I\times R\$, so the power loss is also \$V^2\over R\$.
So did we actually did worse by increasing the voltage ?
What does it mean to dissipate power \$I^2\times R\$ ?
- It first means that the cable, by its nature, resist to the flow of electrons. its electrons like to be in a state of equilibrium and don't like to be pushed by new entrants
- It also says that if, quite naturally, we say that the dissipated power is = the amount flowing, \$I\$ * some force \$F\$ to be overcome, then that force is proportional to the amount flowing itself : the more there is, the more the force will be strong. Of course, we can give that force a name, and that is precisely the voltage between the cable extremities.
When you think of it, it is not surprising that the dissipated power is quadratic.
If you have very big cable, then it would make sense that dissipated power is linear. You pay a constant price for each electron that comes in. In a smaller cable, the cable becomes saturated and its capacity to accept new electron diminishes.
Putting it all together
Having said all that, it is quite clear what the error of the naive reasoning is : we were using the voltage between the ground and the first extremity of the cable. but the only quantity that makes sense is the voltage across endpoints of the cable.
Another view on this, is that every time you speak of a voltage, you have to know not only the amount of Volt it has but also the 2 points it refers to. They are part of the definition. In itself, a tension of 10 Volts has no physical meaning.
A tension of 10 Volts between point A and point B, on the contrary, does have a meaning.
Going back to the problem, by increasing the voltage between the ground and the 1st extremity of the cable, we need a lower intensity to transmit the same amount of energy to someone else, who will take take this current and consume it at ground level voltage.
Conclusion
That lower intensity yields to a lower dissipated energy \$I^2\times R = I\times V_2\$ in the cable of resistance \$R\$, where \$V_2 = I\times R\$ is the voltage drop across the cable.
An equivalent way of seeing this is that it will induce a lower voltage drop between the central and the consumer.
The limit is that you need to have special equipment. At one extreme if the tension so too high, the electron of the air itself will be pushed around, and electrical discharge (aka a "plasma") will be created.