The answer is $3$. This can be obtained by asking $(10,0,0,0)$, $(0,10,0,0)$, $(0,0,10,0)$. The first time, Ana will tell her $|a_1-10|+|a_2-0|+|a_3-0|+|a_4-0|= 10-a_1+a_2+a_3+a_4 = 20-2a_1$, whence Banana can obtain $a_1$. Similarly, she can get $a_2$ and $a_3$, whence $a_4=10-a_1-a_2-a_3$. Now, we will show optimality. Suppose that Banana could guarantee to find the quadruple in two moves for the sake of contradiction. We know that each value Ana tells Banana must be even and from $0$ to $20$, inclusive: \[|a_1-x_1|+|a_2-x_2|+|a_3-x_3|+|a_4-x_4|\pmod 2 \equiv a_1+a_2+a_3+a_4-x_1-x_2-x_3-x_4\pmod 2 \equiv 0.\] If we view Banana's decision process as a "decision tree," then there are at most $11$ choices for how she could ask her second question (based on the result of the first question, since we assumed that such a strategy exists). Each of these questions could have at most $11$ different answers from Ana. Thus, the total number of possible cases that Banana could determine is at most $11^2$. However, there are $\binom{10+3}{3} = 286$ possible tuples for Ana to start with, so Banana cannot determine all of the tuples that Ana could have chosen, a contradiction.

Quite similar to ISL 2002 C4; the bound part (checking the number of possibilities) is the same. https://artofproblemsolving.com/community/c6h17339p118715