= Solution
Write $\mathrm{Me=CH_3}$. The alternating-vertex cubane A is $\mathrm{Li_4Me_4}$; replacing one or two methyl vertices gives $\mathrm B=\mathrm{Li_4Me_3X}$ and $\mathrm C=\mathrm{Li_4Me_2X_2}$. Balanced redistributions are
$$
\mathrm{A+LiX\rightleftharpoons B+MeLi},\qquad
\mathrm{A+2LiX\rightleftharpoons C+2MeLi}.
$$
The smaller aggregate has composition $\mathrm D=\mathrm{Li_2MeX}$, since the conventional masses $2(7)+15+80=109$ or $2(7)+15+127=156$ match the supplied values. Thus $\mathrm{A+4LiX\rightleftharpoons4D}$ is another balanced redistribution.
For the structures, place Li on one tetrahedral set of cube corners and Me/X on the complementary set, joining adjacent unlike vertices. A contains four Me vertices, B three Me and one X, and C two of each. D is a four-membered alternating Li-Me-Li-X ring. The quoted relative masses use conventional elemental masses; a fully $^6$Li-enriched D would be approximately two mass units lighter, without changing its composition.
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