We claim that the answer is $\boxed{n = 6}$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.624792061418095cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.054879523111556, xmax = 8.194704599724632, ymin = -3.9999826215199046, ymax = 3.8616261271371033; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-4.35,1.65)--(-5.2883225910085745,-2.101866013632178)--(-4.791796534757198,-1.2776327602548967)--(-1.69,-1.33)--(-2.954650126627112,-0.9201339997539073)--(-4.622977675631731,2.575409436255767)--cycle, linewidth(2.) + rvwvcq); /* draw figures */ draw((-4.35,1.65)--(-5.2883225910085745,-2.101866013632178), linewidth(0.8) + rvwvcq); draw((-5.2883225910085745,-2.101866013632178)--(-4.791796534757198,-1.2776327602548967), linewidth(0.8) + rvwvcq); draw((-4.791796534757198,-1.2776327602548967)--(-1.69,-1.33), linewidth(0.8) + rvwvcq); draw((-1.69,-1.33)--(-2.954650126627112,-0.9201339997539073), linewidth(0.8) + rvwvcq); draw((-2.954650126627112,-0.9201339997539073)--(-4.622977675631731,2.575409436255767), linewidth(0.8) + rvwvcq); draw((-4.622977675631731,2.575409436255767)--(-4.35,1.65), linewidth(0.8) + rvwvcq); draw((-4.35,1.65)--(-3.4798720847058426,-1.2997818153052403), linewidth(0.8) + wrwrwr); draw((-4.791796534757198,-1.2776327602548967)--(-3.6715592036863796,0.581961209322653), linewidth(0.8) + wrwrwr); draw((-2.954650126627112,-0.9201339997539073)--(-4.839964490390894,-0.3091142076290535), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((-4.35,1.65),linewidth(2.pt) + dotstyle); dot((-5.2883225910085745,-2.101866013632178),linewidth(2.pt) + dotstyle); dot((-4.791796534757198,-1.2776327602548967),linewidth(2.pt) + dotstyle); dot((-1.69,-1.33),linewidth(2.pt) + dotstyle); dot((-2.954650126627112,-0.9201339997539073),linewidth(2.pt) + dotstyle); dot((-4.622977675631731,2.575409436255767),linewidth(2.pt) + dotstyle); dot((-3.947702239129861,-0.4037469012524781),linewidth(2.pt) + dotstyle); label("$P$", (-4.01622840445246,-0.3071601404179702), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] For $n = 6$, the above polygon, with a lightbulb placed at $P$ works. For $n= 3$, a lightbulb placed at any point inside or on the triangle lights up all three sides of the triangle. For $n = 4$, draw the diagonal contained inside the quadrilateral. Any lightbulb placed will be inside one of the two triangles thus formed, hence will fully light up at least two sides of the quadrilateral. For $n = 5$, again we take a diagonal which is fully contained inside the polygon. This separates the polygon into a triangle and a quadrilateral. If the lightbulb is placed inside the triangle it will light up the sides of the polygon which are part of the triangle, and if it is placed on the quadrilateral, it will light up two sides of the quadrilateral, and hence will light up at least one side of the polygon. This finishes the first part. $\spadesuit$ For the second part, we may simply recall the Art Gallery Problem which immediately gives us the desired. Recalling the proof, it goes by triangulating the polygon and 3-coloring the vertices. For an alternate way, one may do casework to prove that a polygon with $n \le 5$ can be fully lightened by a single lightbulb, and then argue that if we have a reflex angle, then we may divide the hexagon into two polygons with $n \le 5$ by extending one of the arms of the reflex angle. (If we do not have a reflex angle then it's obvious) $\square$