There are $4!=24$ such numbers. We can index them with $i=1,…,24$ and write each of them as
$n_{i}=a_{i}+10b_{i}+100c_{i}+1000d_{i}$
Then
$\sum_{i=1}^{24}n_{i}=\sum_{i=1}^{24}a_{i}+10\sum_{i=1}^{24}b_{i}+100\sum_{i=1}^{24}c_{i}+1000\sum_{i=1}^{24}d_{i}$
$\left\{ i\mid a_{i}=5\right\}, \left\{ i\mid a_{i}=6\right\},\left\{ i\mid a_{i}=7\right\}$ and $\left\{ i\mid a_{i}=8\right\}$
Since it's a $4$ digit number, each digit will appear $6=\frac{24}{4}$ times in each of units, tens, hundreds, and thousands place, So:
$\sum_{i=1}^{24}a_{i}=6.5+6.6+6.7+6.8=6(5+6+7+8)=156$
This can also be applied for $b,c$ and $d$ and finally we find:
$\sum_{i=1}^{24}n_{i}=156+10*156+100*156+1000*156=1111*156=173316$