Let $d_{i}$ denote the number of cliques of which person $i$ is a member. Clearly $d_{i} \geq 2$. We now distinguish two cases: (i) For some $i, d_{i}=2$. Suppose that $i$ is a member of two cliques, $C_{p}$ and $C_{q}$. Then $\left|C_{p}\right|=\left|C_{q}\right|=n$, since for each couple other than $i$ and his/her spouse, one member is in $C_{p}$ and one in $C_{q}$. There are thus $(n-1)(n-2)$ pairs $(r, s)$ of nonspouse persons distinct from $i,$ where $r \in C_{p}, s \in C_{q} .$ We observe that each such pair accounts for a different clique. Otherwise, we find two members of $C_{p}$ or $C_{q}$ who belong to one other clique. It follows that $k \geq 2+(n-1)(n-2) \geq 2 n$ for $n \geq 4$. (ii) For every $i, d_{i} \geq 3$. Suppose that $k<2 n$. For $i=1,2, \ldots, 2 n$ assign to person $i$ an indeterminant $x_{i}$, and for $j=1,2, \ldots, k$ set $y=\sum_{i \in C_{j}} x_{i} .$ We know that if $k<2 n$, then there exist $x_{1}, x_{2}, \ldots, x_{2 n}$, not all zero, such that $y_{1}=y_{2}=\cdots=y_{k}=0$. On the other hand, suppose that $y_{1}=y_{2}=\cdots=y_{k}=0$. Let $M$ be the set of the couples and $M^{\prime}$ the set of all other pairs of persons. Then $$ \begin{aligned} 0 &=\sum_{j=1}^{k} y_{j}^{2}=\sum_{i=1}^{2 n} d_{i} x_{i}^{2}+2 \sum_{(i, j) \in M^{\prime}} x_{i} x_{j} \\ &=\sum_{i=1}^{2 n}\left(d_{i}-2\right) x_{i}^{2}+\left(x_{1}+x_{2}+\cdots+x_{2 n}\right)^{2}+\sum_{(i, j) \in M}\left(x_{i}-x_{j}\right)^{2} \\ & \geq \sum_{i=1}^{2 n} x_{i}^{2}>0 \end{aligned} $$if not all $x_{1}, x_{2}, \ldots, x_{2 n}$ are zero, which is a contradiction. Hence $k \geq 2 n$. Q.E.D $\textbf{Remark:}$ For $n=3$, there exists a collection of 4 cliques satisfying the conditions $\{(1,2,3),(3,4,5),(5,6,1),(2,4,6)\}$ where the 3 couples are $\{(1,4),(2,5),(3,6)\}$. Thus we need $n \geq 4$.