Question 1108845
There are no other solutions. My proof is below.
The kids at artofproblemsolving would read it and laugh.
They would probably say that it was obvious based on the Ruszcyk theorem,
or some such thing.
(I just made that theorem up, but they probably have a few modular arithmetic theorems).
 
 
Let the numbers be made of the digits {{{a}}} , {{{b}}} , and {{{c}}} .
{{{n=100a+10b+c}}} and {{{m=100c+10b+a}}} are 3-digit numbers, so {{{a<>0}}} and {{{c<>0}}}.
To have {{{n<>m}}} , it must be {{{a<>c}}} ,
and to have {{{n-m>0}}} , it must be {{{a>c}}} .
 
For any two numbers x and y,
if x-y and x+7 are both multiples of 7,
x and y are both multiples of 7,
and vice versa.
So, m and n must be multiples of 7.
For {{{n=100a+10b+c=(7*14+2)a+(7+3)b+c=7(14a+b)+(2a+3b+c)}}} to be a multiple of 7,
{{{2a+3b+c}}} must be a multiple of 7.
Similarly,  for {{{m=100c+10b+a}}} to be a multiple of 7,
{{{2c+3b+a}}} must be a multiple of 7.
If {{{2a+3b+c}}} and {{{2c+3b+a}}} are both multiples of 7,
{{{(2a+3b+c)-(2c+3b+a)=a-c}}} must be a multiple of 7, and
{{{(2a+3b+c)+(2c+3b+a)=3(a+c)+6b=3(a+c+2b)}}} must be a multiple of 7,
which means {{{a+c+2b}}} must be a multiple of 7.
 
For {{{a-c}}} to be a multiple of 7, the only choices are
{{{system(c=1,a=1+7=8)}}} and  {{{system(c=2,a=2+7=9)}}} .
 
With {{{system(c=1,a=8)}}} , {{{a+c=9}}} ,
and then for {{{a+c+2b=9+2b}}} to be a multiple of 7,
it must be {{{9+2b=21}}} , {{{b=6}}} .
So one solution is {{{system(a=8,b=6,c=1)}}} or {{{system(n=861,m=168)}}} .
 
With {{{system(c=2,a=9)}}} , {{{a+c=11}}} ,
and then for {{{a+c+2b=11+2b}}} to be a multiple of 7,
it must be {{{11+2b=21}}} , {{{b=5}}} .
So the other solution is {{{system(a=9,b=5,c=2)}}} or {{{system(n=952,m=259)}}} .