A number theory problem by A Former Brilliant Member

For the thousands' place, we may choose any of the given $4$ digits; for the hundreds' place, anyone of the remaining $3$ ; for the tens' place, either one of the remaining $2$ . The units' digit is assigned the fourth of the given digits. In all, there are $4 \times 3 \times 2 \times 1 = 24$ possibilities.

Each digit appears $6$ times in each place. Therefore, the sum is

$6(5555)+6(4444)+6(3333)+6(2222)=6(15554)=$ $\boxed{93324}$

The complete list of numbers formed is as follows:

$2345,2354,2435,2453,2534,2543,3245,3254,3425,3452,3524,3542,4235,4253,4325,4352,4523,4532$

$5234,5243,5324,5342,5423,5432$