This is definitely very far from being an answer, but still.
Let $f(t)=\sum_nF(n)t^n$ be the ogf for the numbers in question.
As in the comment by Timothy Chow, call a pair $\langle\boldsymbol a,\boldsymbol b\rangle$ indecomposable if $\boldsymbol a$ and $\boldsymbol b$ do not have common partial sums, that is, $a_1+...+a_i\ne b_1+...+b_j$ except for 0 and total. Let $\bar F(n)$ be the number of indecomposable unmatchable pairs with sum $=n$. The sequence (halved) starts with $0,1,3,10,27,80,216,...$
Let $\bar f(t)$ be the generating function for these.
It is clear that each pair decomposes uniquely into indecomposables, hence $f=1+\bar f+\bar f^2+\bar f^3+...=\frac1{1-\bar f}$, so it suffices to compute $\bar f$.
Next, indecomposables can be similarly uniquely "glued" from "elementary" unmatchable pairs. A pair $\langle\boldsymbol a,\boldsymbol b\rangle$ is elementary if either $\boldsymbol a$ or $\boldsymbol b$ consists of a single number.
By a gluing I mean a situation like this: $$\langle(a_1),(b_1,...,b_{j-1},x)\rangle;\langle(x,a_2,...,a_{i-1},y),(b_j)\rangle;\langle(a_i),(y,b_{j+1},...,b_{j'-1},z)\rangle;\langle(z,a_{i+1},...,a_{i'-1},t),(b_{j'})\rangle,...$$
Now it is not difficult to count elementary unmatchable pairs with given "ends". For fixed $x$ and $y$ let $F_{xy}(n)$ be the number of unmatchable pairs of the form $\langle(x+n+y),(x,...,y)\rangle$ and let $f_{xy}(t)=\sum F_{xy}(n)t^n$ be the corresponding generating function; then $$ f_{xy}(t)=\frac{1-3t^2+2t^3-2^{d-1}t^{d+1}+2^dt^{d+3}}{(1-t)(1-4t^2)}-\begin{cases}0,&x\ne y\\\frac t{2(1-t)},&x=y\end{cases} $$ where $d=|x-y|$.
Uniqueness of gluing then gives $$ \bar f(t)=\sum_{k,x_0,x_1,x_2,...,x_k>0}f_{x_0x_1}(t)f_{x_1x_2}(t)\cdots f_{x_{k-1}x_k}(t)t^{x_0+x_1+x_2+...+x_k}. $$ I had some functional equations when making a series of two variables out of this, but could not simplify it any further.