$$A+B = (a+d) \cdot 10^3 + (b+c) \cdot 10^2 + (b+c) \cdot 10 + (a+d).$$
All digits of $A + B$ are odd. However, in order to find the digits of $A + B$ we must know if the integers $a+d$ and $b+c$ are less than $10$. Hence we have the cases:

(α) Let $a+d \ge 10$ and $b+c \ge 10$. Then, because $a > b > c > d > 0$, we have:
$$\begin{aligned} a+d &= 10+k, \quad k = 0,1,2,...,5, \\ b+c &= 10+\ell, \quad \ell = 0,1,2,...,5. \end{aligned}$$
Hence we have:
$$\begin{aligned} A+B &= (10+k) \cdot 10^3 + (10+\ell) \cdot 10^2 + (10+\ell) \cdot 10 + (10+k) \\ &= 10^4 + (k+1) \cdot 10^3 + (\ell+1) \cdot 10^2 + (\ell+1) \cdot 10 + k, \end{aligned}$$
that is $A + B$ has digits $1, k+1, \ell+1, \ell+1, k$, (all must be odd), absurd.

(β) Let $a+d \ge 10$ and $b+c < 10$. Then, since $a > b > c > d > 0$, we have:
$a+d = 10+k, \quad k = 0,1,2,...,5$ and:
$$\begin{aligned} A+B &= (10+k) \cdot 10^3 + (b+c) \cdot 10^2 + (b+c) \cdot 10 + (10+k) \\ &= 10^4 + k \cdot 10^3 + (b+c) \cdot 10^2 + (b+c+1) \cdot 10 + k. \end{aligned}$$
Therefore we have the cases:
* If $b+c=9$, then $A+B$ has the digit of decades $0$, (absurd)
* If $b+c < 9$, then $A+B$ has as digits the integers $b+c$ and $b+c+1$ which is not possible both to be odd.

(γ) Let $a+d<10$ and $b+c \ge 10$. Then, since $a>b>c>d>0$, we have:
$b+c=10+\ell, \quad \ell=0,1,2,...,5$ and $A+B$ is written
$$A+B = (a+d) \cdot 10^3 + (10+\ell) \cdot 10^2 + (10+\ell) \cdot 10 + (a+d) \\ = (a+d+1) \cdot 10^3 + (\ell+1) \cdot 10^2 + \ell \cdot 10 + (a+d),$$
which means that $\ell$ and $\ell+1$ both are odd digits, (absurd)

(δ) Let $a+d<10$ and $b+c<10$. Then $a+d$ and $b+c$, must be odd. Since $a>b>c>d>0$ and $a \ge 7$, it follows that $a+d=9$ and since $5 \ge c \ge 2$, $6 \ge b \ge 3$, it follows $10>b+c \ge 5$, i.e. $b+c \in \{5,7,9\}$
Therefore we have the cases:
$a+d=9$ with $a=8, d=1$ and $b+c=9$ with $b=7, c=2$ or $b=6, c=3$ or $b=5, c=4$
Hence: $A = 8721$ or $A = 8631$ or $A = 8541$.
$a+d=9$ with $a=7, d=2$ and $b+c=9, b=6, c=3$ or $b=5, c=4$.
Hence: $A = 7632$ or $A = 7542$
$a+d=9$ with $a=8, d=1$ and $b+c=7$ with $b=5, c=2$ or $b=4, c=3$.
Hence: $A = 8521$ or $A = 8431$.
$a+d=9$ with $a=7, d=2$ and $b+c=7$ with $b=4, c=3$.
Hence: $A = 7432$.
$a+d=9$ with $a=8, d=1$ and $b+c=5$ with $b=3, c=2$.
Hence: $A = 8321$.