  We present a new variant of the secretary problem. Let $A$ be a totally
ordered set of $n$ \emph{applicants}.
  Given $P\subseteq A$ and $x\in A$, let $rr(P,x)=\vert\{z\in P \mid z\leq
x\}\vert\mbox{ }$ be the \emph{relative rank of} $x$ \emph{with regard to} $P$,
and let $rr_n(x)=rr(A,x)$. Let $x_1,x_2,\dots,x_n\in A$ be a random sequence of
distinct applicants. The aim is to select $1<j\leq n$ such that
$rr_n(x_{j-1})-rr_n(x_j)\in\{-1,1\}$.
  Let $\alpha$ be a real constant with $0<\alpha<1$. Suppose the following
stopping rule $\tau_n(\alpha)$: reject first $\alpha n$ applicants and then
select the first $x_j$ such that $rr(P_j,x_{j-1})-rr(P_j,x_j)\in\{-1,1\}$,
where $P_j=\{x_i\mid 1\leq i\leq j\}$. Let $p_{n,\tau}(\alpha)$ be the
probability that $\tau_n(\alpha)$ selects $x_j$ such that
$rr_n(x_{j-1})-rr_n(x_j)\in\{-1,1\}$. We show that
\[\lim_{n\rightarrow\infty}p_{n,\tau}(\alpha)\leq
\lim_{n\rightarrow\infty}p_{n,\tau}\left(\frac{1}{2}\right)=\frac{1}{2}\mbox{.}\]
