for even n,k where n>2 $K(n,k)=\{n\}$

How? Can you write out the solution.

I think the answer in post #2 is incorrect. We claim that the answer is $2^7 - 1 = \boxed{127}.$ First we will show the following lemma. Lemma. For $m \ge n$, we have that $K(n, m) = K(n, m+1).$ Proof. We will prove this by strong induction on $n$. If $n = 1$, this is easy to check. Suppose it holds for $n \le k.$ When $n = k+1$, fix some $m \ge k+1.$ Notice that for $1 \le i \le k$, $K(i, m-1) = K(i, m)$ from the induction hypothesis. Therefore, it becomes clear from the definition of $K$ that $K(k+1, m) = K(k+1, m+1)$ (since $K(i, m-1) \cap K(k+1-i, m-1) = \emptyset \Leftrightarrow K(i, m) \cap K(k+1-i, m) = \emptyset$). The induction is complete. $\blacksquare$ Inspired by the lemma, let's define the set $S_n = K(n, n)$ for each $n \in \mathbb{N}.$ From the lemma, we then know that $S_n = \left\{ \left. k \right|\text{ }1\le k\le n\text{ and }S_k\cap S_{n-k}=\varnothing \right\}$. After writing out the first few sets of the sequence of sets $\{S_i\}$, we can conjecture the following: Conjecture For positive integers $n$, the set $S_n$ is the set of all integers $m$ so that the unique set of powers of two with sum $m$ is a subset of the unique set of powers of two with sum $n$. For convenience, call the set of powers of two summing to $k$ the tasty set of $k.$ Proof. The proof goes by strong induction on $n$. When $n = 1$ or $2$, this is easily checked. Suppose it holds for $n < k$. When $n = k$, we wish to show that for $1 \le \ell \le k$, $S_{\ell} \cap S_{k - \ell} = \varnothing$ if and only if the tasty set of $\ell$ is a subset of the tasty set of $k.$ Indeed this is easy to check. The if direction is obvious. As for the only if direction, we will instead check the contrapositive. If $\ell$ is an integer so that the tasty set of $\ell$ is not a subset of the tasty set of $k$, then the tasty set of $k - \ell$ must not be disjoint with the tasty set of $\ell$, because if that were the case then the tasty set of $k$ would be the disjoint union of the tasty sets of $\ell, k- \ell$, clearly absurd. This means that there is a common $1$ in the binary representations of $\ell$ and $k- \ell$, and so the number with this binary representation is in both $S_{\ell}$ and $S_{k - \ell}$. Hence $\ell \in S_k$, and the contrapositive is proven. $\blacksquare$ It is easy to enumerate the number of positive integers with tasty sets contained in the tasty set of $2004$ as $\boxed{127}.$ By the Conjecture, that is our answer. $\square$