Three, the two postal codes on an envelope
$M$ and $N$ are both composed of the six different digits $0,1,2,3,5,6$. There are four codes as follows:
A: 320651,
B: 105263,
$C: \quad 612305$,
D: 316250.
It is known that codes $A, B, C, D$ each have exactly two digits in the same positions as $M$ and $N$. $D$ has exactly three digits in the same positions as $M$ and $N$. Try to find: $M$ and $N$.
For the code $M$, consider the code $A$ where exactly two digits are the same as the corresponding digits in $M$. Let these two positions be $x_{1}, x_{2}$. Since the digits in these two positions in $B$ and $C$ are different from those in $A$, the digits in these two positions in $B$ and $C$ must also be different from those in $M$. Therefore, the positions in $B$ where the digits are the same as those in $M$ must be different from $x_{1}, x_{2}$. Let these positions be $x_{3}, x_{4}$; similarly, the positions in $C$ where the digits are the same as those in $M$ can only be $x_{5}, x_{6}$, which are different from $x_{1}, x_{2}, x_{3}, x_{4}$. The same conclusion applies to $N$. This means that for each digit position, among the digits of $A, B, C$ at that position, one must be the same as the digit of $M$ at that position; similarly, one must be the same as the digit of $N$ at that position.

From this, we know that the digits 6 and 0 in $D$ cannot be the digits of $M$ and $N$ at the corresponding positions. Therefore, among the digits 3, 1, 2, 5 in $D$, only one digit is different from the digit of $M$ at the corresponding position; the same applies when comparing with $N$.
(1) If 3 is incorrect, then we have 610253, 013256;
(2) If 1 is incorrect, then we have 360251, 301256;
(3) If 2 is incorrect, then we have 312056, 310652;
(4) If 5 is incorrect, then we have 310265, 315206.
Upon inspection, it is found that the codes $M, N$ on the envelope are either both 610253, or both 310265, or one is 610253 and the other is 310265.