The thousands digit is 3, and the hundredths digit is 6.
Let the tens digit be a, and ones digit be b

3+6+a=b or 3+6+b=a or 3+a+b=6 or 6+a+b=3

That is,

9+a=b, 9+b=a, a+b=3, or a+b=-3

The last one is out, so we have

9+a=b, 9+b=a, a+b=3
 
The only solution for 9+a=b is a=0, b=9, which gives 3609

The only solution for 9+b=a is b=0, a=9, which gives 3690

The solution for a+b=3 are

 a=0, b=3, which gives 3603
 a=1, b=2, which gives 3612
 a=2, b=1, which gives 3621
 a=3, b=0, which gives 3630

So all 6 solutions are:

1.  3603
2.  3609
3.  3612
4.  3621
5.  3630
6.  3690

Edwin