Conjecture on the sum-freedom of the multiplicative inverse function

Let $\mathbb F_{2^n}$ be the finite field with $2^n$ elements, and define the multiplicative inverse function $F_{\operatorname{inv}}:\mathbb F_{2^n}\to\mathbb F_{2^n}$ by

$$F_{\operatorname{inv}}(x)=x^{2^n-2}=\begin{cases}x^{-1}&\text{if }x\ne 0,\\0&\text{if }x=0. \end{cases}$$

For an integer $k$, say that $F_{\operatorname{inv}}$ is $k$th order sum-free when the sum of its values over every $k$-dimensional affine subspace is nonzero; equivalently, it is not $k$th order sum-free precisely when there is a $k$-dimensional linear subspace $E$ of $\mathbb F_{2^n}$ such that

$$\sum_{0\ne x\in E}\frac{1}{x}=0.$$

Sum-freedom conjecture for the multiplicative inverse function. For even $n$, $F_{\operatorname{inv}}$ is not $k$th order sum-free for $2\le k\le n-2$. For odd $n$, $F_{\operatorname{inv}}$ is not $k$th order sum-free for $3\le k\le n-3$.

Known results establish several cases, including non-sum-freedom when $\gcd(k,n)>1$, symmetry between orders $k$ and $n-k$, and the cases forced by divisibility conditions; computer investigations suggest that the displayed parity-dependent pattern gives the complete answer.