Solution:

The sum of $20$ has $12$ answers, and this is the largest number of answers for any four-digit clue.

We could simply list all the possible sets of four digits and then count. There are $126$ such sets.

Alternatively, define $A(s, n, k)$ to be the number of options with sum $s$ using exactly $n$ digits whose largest digit is less than or equal to $k$. Then the question is to find the maximum of $A(s, 4,9)$ for all values of $s$.

By symmetry, since we can replace each digit $d$ with $10-d$, we know that $A(s, 4,9)=A(40-s, 4,9)$, so we only need to investigate values of $s$ from the minimum, $1+2+3+4=10$, through $20$. (This also implies we only need to list the $69$ sets of four digits whose sum is less than or equal to $20$ in order to prove that $20$ has the most answers. In fact, we can use even fewer than that, since by adding $1$ to the largest digit we can see that $A(s, n, 9) \leq A(s+1, n, 9)$ as long as there are no ways of writing $s$ using the digit $9$; for sums of four digits this shows we only need to investigate sums of $15$ through $20$.)

To compute $A(s, n, k)$ in general, we note that any sum must either use a digit equal to $k$ or not. If there is a digit equal to $k$, then there are $A(s-k, n-1, k-1)$ ways to finish the sum. If there is no digit $k$, then there are $A(s, n, k-1)$ ways to finish the sum. Thus, $A(s, n, k)=A(s-k, n-1, k-1)+A(s, n, k-1)$.

We also know that $A(s, n, k)$ is $0$ in a lot of cases, including any where $k<n$, and $A(s, 1, k)$ is equal to $1$ when $0<s<k$ and $0$ otherwise, because we must have one digit that equals $s$.

Thus, we can fill in the following tables, beginning with $n=2$, and then $n=3$, and then finally $n=4$.

We see that $20$ has $12$ answers, while $19$ and $18$ have only $11$ answers (and similarly $21$ and $22$ also have $11$ answers), and the remaining numbers have even fewer answers.