Couldn't This Wait Till Next Year?

The answer is $1^2 + 2^2 + \cdots + 2015^2.$ I don't have a complete solution to this problem yet; here's a construction which shows this is the minimum. This problem is taken from China TST 2013. Also, it is getting pretty late here so here's just an outline.

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Label the persons $1, 2, \cdots , 2015.$ Let $C_i$ be the number of cards person $i$ has. After $n$ moves, consider $\triangle _n= \displaystyle \sum_{i=1} ^{2015} C_i \cdot i.$ Suppose, person $j$ gives a card to person $j+1$ on the $(n+1)$ th move, so $\triangle _{n+1} - \triangle _n= (C_j-1) \cdot j + (C_{j+1} + 1) \cdot (j+1) -C_j \cdot j - C_{j+1} \cdot (j+1)=1,$ so $\triangle_n$ increases by $1.$ If person $j$ gives a card to person $j-1,$ similarly we can show that $\triangle_n$ decreases by $1.$ Note that when all persons have the same number of cards, for all $j,$ $C_j= \dfrac{1 + 2 + \cdots + 2015}{2015} = \dfrac{2016}{2},$ so $\triangle= (1 + 2 + \cdots + 2015) \dfrac{2016}{2} = \dfrac{2015 \times 2016^2}{4}.$ Set $a_i= 2i$ and $a_{i+1007} = 2(1007-i)^2-1$ for all $i \leq 1008.$ With this configuration we can easily verify that $\triangle _0 = \dfrac{2015 \times 2016^2}{4} + (1^2 + 2^2 + \cdots + 2015^2).$ Since $\triangle_n$ decreases by at most $1$ each move, it will take at least $1^2 + 2^2 + \cdots + 2015^2$ moves to reach there.

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I'm still working on proving that this is indeed the minimum. Any help will be appreciated. Also, since I'm feeling pretty sleepy right now and my notebook isn't here with me, the construction for $a_i$ might not be correct. If that is the case, I'll fix it soon.