China Girls' Mathematical Olympiad

The answer is $16n-8$. Assume that $P_i=(x_i,y_i)$ for $1\le i\le 4n+1$. Set $$P_{4n+2}=(x_{4n+2},y_{4n+2})=P_1=(x_1,y_1).$$ We will show that the sum $$S=\sum _{i=1}^{4n+1}(P_i{P}_{i+1}{)}^2\ge 16n-8.$$ First, we show that this minimum can be obtained by setting $$(P_1,P_2,\dots ,P_{4n+1})=((2,0),(4,0),\dots ,(n,0),(n-1,0),\dots ,(1,0),(0,2),(0,4),\dots ,(0,n),(0,n-1),\dots ,(0,1),(-2,0),(-4,0),\dots ,(-n,0),(-n+1,0),\dots ,(-1,0),(0,-2),(0,-4),\dots ,(0,-n),(0,-n+1),\dots ,(0,-1),(0,0))$$ for $n$ even, and $$(P_1,P_2,\dots ,P_{4n+1})=((1,0),(3,0),\dots ,(n,0),(n-1,0),\dots ,(2,0),(0,1),(0,3),\dots ,(0,n),(0,n-1),\dots ,(0,2),(-1,0),(-3,0),\dots ,(-n,0),(-n+1,0),\dots ,(-2,0),(0,-1),(0,-3),\dots ,(0,-n),(0,-n+1),\dots ,(0,-2),(0,0))$$ for $n$ odd. This can be easily checked. For example, when $n=2m$ is even, our construction shows that $$\begin{array}{rl}S& =4\left(\sum _{i=1}^{m-1}(P_i{P}_{i+1}{)}^2+(P_m{P}_{m+1}{)}^2+\sum _{i=m+1}^{2m-1}(P_i{P}_{i+1}{)}^2\right)\\ & +3(P_{2m}{P}_{2m+1}{)}^2+(P_{4n}{P}_{4n+1}{)}^2+(P_{4n+1}{P}_1{)}^2\\ & =4(4(m-1)+1+4(m-1))+3\times 5+1+4\\ & =32m-8=16n-8.\end{array}$$ The exact same argument works when $n$ is odd. Note that $$S=\sum _{i=1}^{4n+1}[(x_i-x_{i+1}{)}^2+(y_i-y_{i+1}{)}^2].$$