130714-1128 EDT
The original question is not clear.
If there are no loads on this circuit except at the far end of the 325 ft run, then it makes no sense to change wire size thru the run.
Determine the peak load current at the 325 ft point, and the maximum voltage drop that is desired at this peak load. That deterimines the maximum impedance of the circuit. You can use the impedance as the resistance and then calculatre wire size.
How do you determine a reasonable voltage drop. That depends upon the type of load, and other criteria.
Suppose the load is a motor, then you are more likely to be concerned with inrush current to the motor and the motor's startup time. My DeWalt radial arm saw for example. It does very poorly with a 12" blade, a 120 V supply, and 125 ft run of #12, and a 20 A QO breaker. It can be started once, then a considerable delay before starting to allow the QO to cool down. Try to start twice within 10 to 20 seconds and the breaker trips. Startup time is many seconds, and about 80 A. If the saw is at the end of 20 ft of #12 the startup is a fraction of a second. Thus very much less startup heating of the QO. Usually source voltage is more like 124 V.
If your nominal voltage is low 105 to 110 V, then you have greater problems than if nominal is more in the range of 120 tp 135 V.
Your loop length is 650 ft or about 2/3 of 1000 ft. At a wire temperature of 20 C (68 F) my copper tables have #4 at 0.249, and #8 at 0.628 ohms/1000 ft. Doing an equal split the the total resistance is (0.249*325/1000) + (0.628*325/1000) = 0.285 ohms. Using #6 the whole distance the resistance is 0.395*650/1000 = 0.256 ohms. The total copper weight of each approach is (126.4*325/1000) + (50.0*325/1000) = 57.3 #, and 79.5*650/1000 = 51.7 #.
In this application why would you consider the two stage approach?
Assume the loop resistance is 0.25 ohms and the load current is 100 A, then voltage drop is 25 V. Is this acceptable? Load current was not clearly defined. So now assume the load is never more than 20 A, then maximum voltage drop will be greater than 5 V by a little because the wire will be hotter than 20 C.
.