The sum of the angles of an $n$-gon is $(n-2) 180$, so the average angle measure is $(n-2) 180 / n$. The common difference in this arithmetic progression is at least 1 , so the difference between the largest and smallest angles is at least $n-1$. So the largest angle is at least $(n-1) / 2+(n-2) 180 / n$. Since the polygon is convex, this quantity is no larger than 179: $(n-1) / 2-360 / n \leq-1$, so that $360 / n-n / 2 \geq 1 / 2$. Multiplying by $2 n$ gives $720-n^{2} \geq n$. So $n(n+1) \leq 720$, which forces $n \leq 26$. Of course, since the common difference is an integer, and the angle measures are integers, $(n-2) 180 / n$ must be an integer or a half integer, so $(n-2) 360 / n=360-720 / n$ is an integer, and then $720 / n$ must be an integer. This leaves only $n=3,4,5,6,8,9,10,12,15,16,18,20,24$ as possibilities. When $n$ is even, $(n-2) 180 / n$ is not an angle of the polygon, but the mean of the two middle angles. So the common difference is at least 2 when $(n-2) 180 / n$ is an integer. For $n=20$, the middle angle is 162 , so the largest angle is at least $162+38 / 2=181$, since 38 is no larger than the difference between the smallest and largest angles. For $n=24$, the middle angle is 165 , again leading to a contradiction. So no solution exists for $n=20,24$. All of the others possess solutions: $$\text{n angles 3 59,60,61 4 87,89,91,93 5 106,107,108,109,110 6 115,117,119,121,123,125 8 128,130,132,134,136,138,140,142 9 136, , 144 10 135,137,139, , 153 12 139,141,143, , 161 15 149,150, , 163 16 150,151, , 165 18 143,145, , 177}$$ (These solutions are quite easy to construct.) The desired value is then $3+4+5+6+$ $8+9+10+12+15+16+18=106$.