We have

$\overline{abcd} + \overline{badc} = 16665$

Write two numbers on the left hand side in decimal form we obtain

$\overline{abcd} = 1000a + 100b + 10c + d$

$\overline{badc} = 1000b + 100a + 10d + c$

Hence the left hand side becomes

$(1000a + 100b + 10c + d) + (1000b + 100a + 10d + c) = 16665$

Grouping the similar terms we have

$<-> (1000a + 100a) + (1000b + 100b) + (10c + c) + (10d + d) = 16665$

$<-> 1100a + 1100b + 11c + 11d = 16665$

Take 11 as the common divisor

$<-> 11(100a + 100b + c + d) = 16665$

Devide both sides by 11

$<-> 100a + 100b + c + d = 1515$

$<-> 100(a + b) + (c + d) = 1500 + 15$

$<-> 100(a+b) - 15 \times 100 + (c + d) - 15 = 0$

$<-> 100(a + b - 15) + (c + d - 15) = 0$

Since $a, b, c, d$ are 1-digit numbers, hence $0 < a + b \leq 18$ and $0 \leq c + d \leq 18$.

Deduce that $0 \leq a + b \leq 3$ and $0 \leq c + d \leq 18$.

So the left hand side is greater or equal to zero.

The equality occurs only when $a + b -15 = 0$ an $c + d - 15 = 0$

Hence $a + b = c + d = 15$.

So $a + b + c + d = 30$.