Write $a,b,c$ for the sums of each of the three rows. We have $a+b+c=1+\cdots+9=45$. The fact that each row is a multiple of 9 means that $a,b,c\in\{9,18,27,36\}$. So $a,b,c$ are either $27,9,9$ or $18,18,9$, in some order. As $1+2+3+4=10$, the 4-digit term cannot be $9$, so it is $18$ or $27$. But $27$ is not the sum of four consecutive integers, so the third row is $3,4,5,6$ in some order.
We are left with $1,2,7,8,9$ for the other two rows, one adding to $18$ and the other $9$. We cannot reach $18$ with two digits at all or with three digits not using the $9$, so the first row is necessarily one of the following two possibilities: $9,8,1$ in some order, and the middle one $2,7$; or $2,7,9$ in some order and $1,8$. This last case is easily discarded since the only 4-digit product is $279\times18=5022$, which produces the wrong last row.
Looking at the first row, the last digit cannot be $1$ because that would imply a repeated digit. $2$ and $9$ do not fit, because they would put an $8$ in the last row. Actually, if the first row starts with $8$ or $9$, as $819\times27$ already has five digits, we learn that the first number starts with $1$ and that the second one is $27$.
We have $189\times27=5103$, not good, so it has to be $198\times 27=5346$. So the answer to the question is $8+7+6=21$.
Your conjecture is true. If it isn't, let us suppose that there is a number $m>9$ of $n$ digits such that $dp(m)=m$. As you have checked and also proved by @ChristopherMarley in the comments, $n>2$.
Let's chop off the last digit of $m$ to get $m'$ of $n-1$ digits. Now, $m' \leq m/10$ (since we're taking the floor of $m/10$) and $dp(m') \geq m/9$(since the last digit is $9$ at most) $> m/10$.
Hence, $m' \leq dp(m')$. Continuing this chopping off of last digit, we'll get a number $m_0$ of $2$ digits such that $m_0 \leq dp(m_0)$ but we know that's not possible.
So, such an $m$ doesn't exist i.e. your conjecture is true that single digit numbers are the only numbers with $dp(m)=m$.
Best Answer
The smallest such multiple is 1122222222.
Claim: The number we are looking for has at least ten digits.
Proof: Assume that there is a counterexample with at most nine digits. Then the sum of digits is at most 18, and 18 is only attained by the example 222222222, which is not divisible by 11. So the sum of digits must be 9 in a counterexample. In particular, the sum of digits with alternating signs is between -9 and 9, and it is divisible by 11. So the sum of digits with alternating signs is 0. But this is a contradiction $\pmod 2$: the parity of the sum and the sum with alternating signs should coincide.
So we need 10 digits. The same argument as above shows that the sum of digits cannot be 9. But it is at most 20, so the sum of digits is 18. Again, by the same argument as above, the sum of digits with alternating signs is 0 (between -18 and 18, this time it is even, and divisible by 11).
If there is no digit 1, then there are nine 2's. But then the sum with alternating signs has a different number of 2's with plus and minus sign. So digit 1 is used, and then there must be at least two of them to obtain an even sum. As the sum of digits is 18, there must be two 1's and eight 2's. The smallest number that we can produce from these digits is 1122222222, which is divisible by 99.