Let $P, Q$ be the projections of $E$ on the lines $BD, AC$ By Simpson we have that each of the following triples of point are collinear: $K,P,L; M,N,P; L,Q,M; K,N,Q$ We also have that $PNEKD$ and $NQCME$ are cyclic Then $MN\perp LK \Leftrightarrow PN\perp PL \Leftrightarrow \angle DPK = \angle NPE \Leftrightarrow \angle DEK = \angle NDE$ $\Leftrightarrow DC\parallel EK \Leftrightarrow AD\perp DC \Leftrightarrow AC$ is a diamter Analogously $KN\perp LM \Leftrightarrow BD$ is a diameters. Then $N$ is the orthocentere of $KLM$ $\Leftrightarrow$ $AC$ and $BD$ are diameter $\Leftrightarrow$ $ABCD$ is a rectangle Therefore if $N$ is the orthocentre of $KLM$ for some point $E$ then $ABCD$ is a rectangle, and then $N$ is orthocentre of $KLM$ for any point $E$