The case is true when $ABCD$ is a parallelogram.
Denote the coordinate of each uppercase letter by a lowercase letter.
Then
$k=\frac{a+b}{2}$
$l=\frac{b+c}{2}$
$m=\frac{c+d}{2}$
$n=\frac{a+d}{2}$
If $H_1,H_2,H_3,H_4$ denote the orthocenters of triangles $AKN,BKL,CLM,DNM$ respectively.
Then
$h_1=2a+\frac{b+d}{2}$
$h_2=2b+\frac{a+c}{2}$
$h_3=2c+\frac{b+d}{2}$
$h_4=2d+\frac{a+c}{2}$
Hence $H_1,H_2,H_3,H_4$ form a parallelogram
$\Leftrightarrow h_1+h_3=h_2+h_4$
$\Leftrightarrow 2a+2c+b+d=2b+2d+a+c$
$\Leftrightarrow a+c=b+d$
$\Leftrightarrow$ ABCD is a parallelogram.
Maths is the doctor of science.