Minimalist example of voter profiles yielding different outcomes
First, let me state that this is a beautiful problem!

Here is a proof that $n=4$. To prove that $n \leq 4$, consider the following example:


$m+4$ voters have preference $a \succ d \succ b \succ c$
$m+3$ voters have preference $b \succ d \succ c \succ a$
$m+2$ voters have preference $c \succ d \succ b \succ a$
$m$ voters have preference $d \succ c \succ a \succ b$


where $m \geq 2$. I leave it to you to check that $a$ is selected by the plurality rule, $b$ by the simple run-off, $c$ by the elimination rule and $d$ by the Borda procedure.

Now, let us prove that $n&gt;3$. The cases $n=1$ and $n=2$ are trivial, so let us assume $n=3$ and write $v_1,v_2,v_3$ for the number of voters in  the three voter profiles. Without loss of generality, let us assume that $v_1 \geq v_2 \geq v_3$, that $a$ is selected by the plurality rule, that $b$ is selected by the run-off election, that $c$ is selected by the elimination rule and that $d$ is selected by the Borda count.

The outcome of the plurality rule shows that $a$ is most preferred by voters of group 1 (since it wins with no ties). If $a$ were also preferred by voters of group 2, $a$ would win the run-off election, which is a contradiction. Thus $a$ is most preferred by voters of group 1, $b$ is most preferred by voters of group 2, and $v_1&gt;v_2 \geq v_3$, the latter inequality being strict if $b$ is not most preferred by voters of group 3. But then it is clear that both $a$ and $b$ survive the first two rounds of the elimination rule, i.e. that $c$ and $d$ are eliminated, which is a contradiction.