We can think of it in this way:
We find all cases for the equation without restriction of $A_1\leq12$ and $A_4\leq10$. Then we fill in these cases with a 2-circle Venn Diagram, the first circle ($P$) with cases satisfying $A_1>12$ (yes, $>$), and the second circle ($Q$) with cases satisfying $A_4>10$. Then what we want to find is $P'\cap Q'$ i.e. $\xi - P - Q + P\cap Q$.
Now we have to solve for
$$A_1 + A_2 + A_3 + A_4 = 25\ \fbox*$$
when $\fbox1\ A_1>12\ (A_1\geq13)$, when $\fbox2\ A_4>10\ (A_4\geq10)$, and when $\fbox3$ both occurs.
From now, we think of those 4 "$A$"s as bins, and numbers as "balls".
For $\fbox*$
Using stars and bars we get ${28\choose3}=3276$
Case $\fbox1$
Then it's easy to eliminate the restriction. We fill in box $A_1$ with 13 balls, and then we get an equation we need to solve for (with no restrictions!):
$$A_1 + A_2 + A_3 + A_4 = 12$$
Using stars and bars the answer is ${15\choose3}=455$
Case $\fbox2$
Similarly we get
$$A_1 + A_2 + A_3 + A_4 = 14$$
and hence we get ${17\choose3}=680$
Case $\fbox3$
Similarly we get
$$A_1 + A_2 + A_3 + A_4 = 1$$
and hence we get ${4\choose3}=4$
Hence we have $3276-455-680+4=2145$ which is what we want.$\ \square$
The sum of the digits $1$ through $9$ is odd. They contribute to the parity of the digit sum of the result no matter which row they’re in. The digit sum of the result is odd. Thus there must be an even number of borrowings.
A column that causes borrowing must have a $7$, $8$ or $9$ in the bottom row, so we cannot have four borrowings.
On the other hand, if there were no borrowing at all, the possible pairs in a column would be $9-6-3$, $8-5-2$ and $7-4-1$, but we can use at most one from each of these three groups.
It follows that there are exactly two borrowings. Thus the difference between the digit sums of the rows must be $5\cdot3-2\cdot9=-3$, and since the sum of all digits is $\frac{9(9+1)}2=45$, the top row must sum to $21$ and the bottom row to $24$.
We need to have exactly two of $7$, $8$ and $9$ in the bottom row to cause the two borrowings.
It can’t be $7$ and $8$ because then $7$ would have to be subtracted from $1$ and $8$ from $2$, so the two borrowing columns would have to be the two lending columns.
If it were $8$ and $9$, that would leave a sum of $7$ for the bottom row, so that could be $3,4$ or $2,5$ or $1,6$. It can’t be $3,4$ because one of those needs to be $A_1$; it can’t be $2,5$ because $5$ would need to be subtracted from $8$ or $9$; and it can’t be $1,6$ because $6$ would need to be subtracted from $9$.
Thus $7$ and $9$ are in the bottom row. That leaves a sum of $8$ for the bottom row, which could be $3,5$ or $2,6$. But it can’t be $2,6$, again because $6$ would need to be subtracted from $9$.
Thus we have $3,5,7,9$ in the bottom row and $1,2,4,6,8$ in the top row. So $4$ must be $A_1$, $7$ must be subtracted from $1$, $9$ from $2$, $3$ from $6$ and $5$ from $8$. Thus the lenders must be $4$ and $1$, so the top row must start $412$. That leaves two possibilities for the order of the last two columns, so there are two solutions:
41286 41268
-7953 and -7935
----- -----
33333 33333
The solutions are confirmed by this Java code. (Full disclosure: I initially made a mistake in the proof and wrote the code to find it, so I knew the solution before I completed the proof.)
Best Answer
Hint: Imagine that we have $13$ "digits," the usual $0$ to $9$, and in addition the digits $\alpha$, $\beta$, $\gamma$, to be thought of as representing $10$, $11$, and $12$ respectively.
Pick a strictly descending sequence $x_1, x_2, x_3, x_4$ of length $4$ from this new digit set of $13$ digits. You know how to count how many such strictly descending sequences there are.
Let $a_4=x_4$, $a_3=x_3-1$, $a_2=x_2-2$, $a_1=x_1-3$. Show that this produces every non-increasing sequence $a_1\ge a_2 \ge a_3 \ge a_4$ in one and only one way.
Now we need to adjust for the unpleasant fact that the initial digit of a $4$-digit number cannot be $0$. That translates into the fact that the strictly descending sequence $x_1=3$, $x_2=2$, $x_3=1$, $x_4=0$ is not allowed.
Remark: The above procedure has nothing much to do with digits. The number of never increasing sequences of length $k$ from the set $\{1,2,\dots,n\}$ is the same as the number of strictly decreasing sequences of length $k$ from the set $\{1,2,\dots,n,n+1,\dots, n+k-1\}$.
Added: The number of strictly decreasing sequences of length $4$ from the digits $0, 1,\dots, 9$ is just $\binom{10}{4}$, the number of ways to choose $4$ distinct numbers. If we use the "digits" $0,1,\dots,9,\alpha,\beta,\gamma$, then the number of strictly decreasing sequences is $\binom{12}{4}$. We must subtract $1$ because $3,2,1,0$ is not allowed.