Solution:
There are $4 \times 3 = 12$ numbers with a given digit of $n$ in the units place. Similarly, there are $12$ with it in the tens place and $12$ with it in the hundreds place. So the sum of the 3 digit numbers is $12 \times 111(a+b+c+d+e)$, where $n=abcde$. So $8668a=332b+1232c+1322d+1331e$. We can easily see that $a=1$ is too small and $a=4$ is too big, so $a=2$ or $3$. Obviously $e$ must be even. $0$ is too small, so $e=2,4,6$ or $8$. Working mod $11$, we see that $0=2b+2d$, so $b+d=11$. Working $\bmod 7$, we see that $2a=3b+6d+e$. Using the $\bmod 11$ result, $b=2, d=9$ or $b=3, d=8$ or $b=4, d=7$ or $b=5, d=6$ or $b=6, d=5$ or $b=7, d=4$ or $b=8, d=3$ or $b=9, d=2$. Putting each of these into the $\bmod 7$ result gives $2a-e=4,1,5,2,6,3,0,4$ mod $7$. So putting $a=2$ and remembering that $e$ must be $2,4,6,8$ and that all digits must be different gives $a, b, d, e = 2,4,7,6$ or $2,7,4,8$ or $2,8,3,4$ as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for $c$. Similarly putting $a=4$, gives $a, b, d, e=3,4,7,8$ or $3,5,6,4$ as the only possibilities. Checking, we find the solution above and no others.