Groupoid cardinality and Egyptian fraction representations of 1 It is well-known that any rational number can be represented using a sum of distinct Egyptian fractions (that is, rational fractions of the form $1/n$ with $n\in\mathbb{N}$). This may be proven by establishing a greedy algorithm that constructs a sequence of such decompositions. For instance, 
\begin{align}
1
&= \frac12+\frac13+\frac16 \\
&= \frac12+\frac13+\frac17+\frac1{42}\\
&= \frac12+\frac13+\frac17+\frac1{43}+\frac1{1806}\\
&= \cdots\\&=\sum_{k=1}^\infty \frac{1}{x_k} \tag{1}
\end{align}
where $x_1=2$ and $x_k = 1+\prod_{j=1}^{k-1}x_j$ for $k>1$ (i.e. Sylvester's sequence A000058).
As noted by Bergner & Walker [1], such Egyptian fractions are relevant in the context of groupoid cardinality. To briefly review: Given a groupoid $G$, the groupoid cardinality is defined [2]  by $|G|=\sum_{[\bullet]\in G}\frac{1}{\#\text{Aut}(\bullet)}$ where $[\bullet]$ is a component of $G$ and $\#\text{Aut}(\bullet)$ is the order of the automorphism group of $\bullet$. If $G$ is a group, then this reduces to $|G| = \frac{1}{\# G}$, and if we form the groupoid $G\coprod H$ as a direct union of groups $G,H$ we have $\textstyle |G\coprod H| =\frac{1}{\#G}+\frac{1}{\#H}$.
Here the link to Egyptian fractions arises: Since every rational number can be represented in terms of Egyptian fractions, so too can every rational fraction be obtained as the cardinality of some groupoid. As a key example, since $|\mathbb{Z}/n|=1/n$ the representations in $(1)$ imply $$1 = |\mathbb{Z}/2\coprod \mathbb{Z}/2|=|\mathbb{Z}/2\coprod \mathbb{Z}/3\coprod \mathbb{Z}/6|=\cdots.$$
But this construction, while valid, is quite artificial. In particular, it is far from obvious to me why each successive groupoid has unit cardinality. This suggests the following question: Is there a natural sequence of groupoids $\{G_k\}$, each of unit cardinality, which generate $(1)$?
References:
[1] John Baez, James Dolan, "From Finite Sets to Feynman Diagrams" (arXiv)
[2] Julia Bergner, Christopher Walker, "Groupoid Cardinality and Egyptian Fractions" (JSTOR)
 A: First a notational issue: you shouldn't write $G$ for the one-object groupoid corresponding to $G$. A much better name for this groupoid is $BG$, or $\text{pt} / G$. 
A natural way to write down a groupoid whose groupoid cardinality is $1$ is to write down a groupoid / homotopy quotient $X/G$ where $|X| = |G|$. In turn, a natural class of such quotients are the adjoint quotients $G/G$ where $G$ acts on itself by conjugation. This is equivalent to the groupoid 
$$\coprod_{[g]} BC_G([g])$$ 
where $C_G(-)$ denotes the centralizer and the coproduct runs over the conjugacy classes of $G$. We get a potential Egyptian fraction representation
$$1 = \sum_{[g]} \frac{1}{|C_G([g])|}$$
if the sizes of the centralizers are distinct. The largest $n$ such that $\frac{1}{n}$ appears in this sum will always come from the conjugacy class of the identity, where $C_G(e) = G$, so $n = |G|$, and everything else will divide $|G|$. This identity is just the usual class equation divided by $|G|$. 
To get the first sum this way we want a group of order $6$ with $3$ conjugacy classes having centralizers of orders $2, 3, 6$, or equivalently $3$ conjugacy classes of sizes $3, 2, 1$. There is a unique such group, namely $D_3$. 
To get the second sum this way we want a group of order $42$ with $4$ conjugacy classes having centralizers of orders $2, 3, 7, 42$, or equivalently $4$ conjugacy classes of sizes $21, 14, 6, 1$. Such a group necessarily has elements of orders $2, 3, 7$, so these must be the non-identity conjugacy classes (in the same order as before, since everything centralizes itself). There are $6$ elements of order $7$, so the Sylow $7$-subgroup $C_7$ is normal, and so by Schur-Zassenhaus our group $G$ must be a semidirect product $C_7 \rtimes H$ where $|H| = 6$. We must have $H = D_3$ or else $G$ has an element of order $6$, so $G$ must be the semidirect product $C_7 \rtimes D_3$. There is a unique nontrivial such semidirect product because the only nontrivial action of $S_3$ on $C_7$ is the one where $\sigma \in S_3$ acts by $\text{sgn}(\sigma)$; unfortunately, this group has an element of order $21$. So in fact no group with the desired condition on conjugacy classes exists. 
Nevertheless, the second sum is still related to some geometry, specifically to the geometry of the (2,3,7) triangle group, or rather to the orbifold quotient of the upper half plane by this group. I don't know about the third sum, though. 
