The code with checksum and reverse

1. If you call the code abcd, the reverse condition compares dcba with abcd.
2. The difference can be written using only the differences d − a and c − b.
3. When using the last digit condition, remember that it does not require equality with the entire sum, but rather equality modulo 10.

Answer: $$3924.$$ Let the code be $abcd$. The number obtained by inverting its figures is $dcba$. How this reverse surpasses the original by 369 units: $$
dcba-abcd=369.

$$ That is to say: $$

(1000d+100c+10b+a)-(1000a+100b+10c+d)=369.

$$ Grouping: $$

999(d-a)+90(c-b)=369.

$$ We divide by 9: $$

111(d-a)+10(c-b)=41.

$$ Since $a,b,c,d$ are figures, the differences $d-a$ and $c-b$ are between $-9$ and $9$. so that $$

111(d-a)+10(c-b)=41,

$$ the only possibility is: $$

d-a=1,\qquad c-b=-7.

$$ Therefore: $$

d=a+1,\qquad c=b-7.

$$ Now we use the condition of the last figure: $$

d\equiv a+b+c\pmod{10}.

$$ Replacing $d=a+1$ and $c=b-7$: $$

a+1\equiv a+b+(b-7)\pmod{10}.

$$ Then: $$

1\equiv 2b-7\pmod{10},

$$ so that $$

2b\equiv 8\pmod{10}.
$$ So $b$ must be congruent to 4 modulo 5. Furthermore, since $c=b-7$ must be a number, $b$ can only be 7, 8, or 9. Of those three values, the only one congruent to 4 modulo 5 is: $$b=9.$$ So: $$c=2.$$ It remains to be used that the code is a multiple of 9. The sum of its figures must be a multiple of 9: $$a+9+2+(a+1)=2a+12.$$ Since $a$ is a figure and $d=a+1$, the only valid possibility is: $$2a+12=18,$$ so $$a=3,$$ So: $$d=4.$$ The code is: $$3924.$$ Check: your figures are different; $3+9+2+4=18$, then it is a multiple of 9; $3+9+2=14$, whose remainder when divided by 10 is 4; its reverse is 4293, and $4293-3924=369$.