Let $m$ is maximal then $m|k+l+n \leq 3m$ Case 1: $k+l+n=3m \to k=l=n=m=4$ and $k+l+m+n=16$ Case 2:$k+l+n=2m \to k=3$ $l+n=2m-3,ln=3m$ $l,n$ are roots of $x^2-(2m-3)x+3m=0$ $D=4m^2-24m+9=(2m-6)^2-27$ is square only for $m=6,10$ and so $k+l+m+n=3m=18,30$ Case 3:$k+l+n=m$ $k=2,l+n=m-2,ln=2m$ $l,n$ are roots of $x^2-(m-2)x+2m=0$ $D=m^2+4-12m=(m-6)^2-32 \to m=12,15$ and $k+l+m+n=2m=24$ or $30$ Answer: $16,18,24,30$