Here is what I think proves that for any i, there is no constant ki satisfying f(n,i)≤nki. That is:

Claim. Let i be a positive integer. Then the function f(n,i) is not polynomially bounded in n.

Proof. First consider the case of i=1. A key observation is that if X={x1,…,xn} contains two subsets A and B such that |A|=|B|, AB, and the sum of A is equal to the sum of B, then X cannot be 1-sum-avoiding since assigning cj as follows violates the condition: cj=2 if xj belongs to A but not to B, cj=0 if xj belongs to B but not to A, and cj=1 if neither holds.

Let m be a positive integer and X be a 1-sum-avoiding set of size 2m. By the above observation, all m-element subsets of X must have distinct sums, and therefore the largest sum must be at least $\binom{2m}{m}$. Therefore, the largest element in X must be at least $\binom{2m}{m}/m>2^{m-1}$, which implies that f(2m,1) > 2m−1. This establishes the claim for i=1.

Now observe that if a set X={x1,…,xn} is i-sum-avoiding, then the set {x1i,…,xni} is 1-sum-avoiding. This means that f(n,1) ≤ f(n,i)i. Since we already know that f(n,1) is not polynomially bounded, f(n,i) is not polynomially bounded in n, either. QED.