The lengths, in order, of four consecutive sides of an equiangular hexagon are 1, 7, 2 and 4 units, respectively. What is the sum of the lengths of the two remaining sides?
Name the vertices of the hexagon so that hexagon $ABCDEF$ has $AB=1$, $BC=7$, $CD=2$, and $DE=4$.  The hexagon is equiangular, so each interior angle measures $180(6-2)/6=120$ degrees.  Extend sides $AB$, $CD$, and $EF$ and call their intersection points $G$, $H$, and $J$ as shown.  The exterior angles of the hexagon each measure $180-120=60$ degrees, so triangles $JDE$, $CBH$, $FGA$, and $JHG$ are all equilateral.  It follows that $JD=DE=4$ units and $CH=CB=7$ units.  Therefore the side length $JH$ of triangle $JGH$ is $4+2+7=13$ units.  Turning to side $HG$, we find that $AF=AG=13-(7+1)=5$ units.  Finally, we solve $JG=JE+EF+FG$ for $EF$ to get $EF=13-(4+5)=4$ units.  The sum of the missing sides is $5+4=\boxed{9}$ units.

[asy]
size(6cm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
dotfactor=4;

pair A=(8,0), B=(7,0), C=7*dir(60), D=9*dir(60), Ep=(13,0)+9*dir(120), F=(13,0)+5*dir(120), G=(13,0), H=(0,0), J=13*dir(60);

pair[] dots = {A, B, C, D, Ep, F};

dot(dots);

draw(A--B--C--D--Ep--F--cycle);

draw(B--H--C,dashed);
draw(D--J--Ep,dashed);
draw(F--G--A,dashed);

label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,NW);
label("$D$",D,NW);
label("$E$",Ep,NE);
label("$F$",F,NE);
label("$G$",G,SE);
label("$H$",H,SW);
label("$J$",J,N);

label("$1$",(A+B)/2,N);
label("$7$",(B+C)/2,NE);
label("$2$",(C+D)/2,SE);
label("$4$",(D+Ep)/2,S);
[/asy]