Gale Shapley - Everyone ends up with second choice of partner
Let us number the men $1,2,3,4$ and the women $a,b,c,d$ (the men propose to the women), with the following preferences:
$$
\begin{align*}
&amp;1\colon b \succ c \succ d \succ a &amp; a\colon 1 \succ 2 \succ 3 \succ 4 \\
&amp;2\colon b \succ a \succ d \succ c &amp; b\colon 3 \succ 4 \succ 1 \succ 2 \\
&amp;3\colon a \succ d \succ c \succ b &amp; c\colon 2 \succ 1 \succ 4 \succ 3 \\
&amp;4\colon a \succ b \succ c \succ d &amp; d\colon 4 \succ 3 \succ 2 \succ 1
\end{align*}
$$
Let us trace the Gale-Shapley algorithm:


Round 1M: $1,2$ propose to $b$; $3,4$ propose to $a$.
Round 1W: $b$ accepts $1$; $a$ accepts $3$.
Round 2M: $2$ proposes to $a$; $4$ proposes to $b$.
Round 2W: $a$ accepts $2$, throwing $3$; $b$ accepts $4$, throwing $1$.
Round 3M: $3$ proposes to $d$; $1$ proposes to $c$.
Round 3W: $d$ accepts $3$; $c$ accepts $1$.


In the end, everyone is married to their second choice.