VMO

Suppose that the tuple $(a_1,a_2,\dots ,a_{2014})$ satisfies the given condition. We investigate two following cases.

Case 1. If there exists a zero in this tuple, it is easy to check that there are at least four zeros. Otherwise, if the tuple has four zeros then it satisfies the given condition.

Case 2. Suppose that $a_i\ne 0$ for all $i$, $1\le i\le 2014$. Denote $a_i=x_i/y_i$ where $x_i,y_i\in {\mathbb{Z}}^{+}$ and $\gcd (x_i,y_i)=1$. It is clear that the tuple $(a_1,\dots ,a_{2014})$ satisfies the given condition if and only if the sequence $(t{a}_1,\dots ,t{a}_{2014})$ is satisfied, where $t$ is the least common multiple of $y_1,\dots ,y_{2014}$. Hence, we can assume that $a_i$'s are nonzero integers.

For any prime $p$, denote $Z_p=(z_1,z_2,\dots ,z_{2014})$ to be the set of exponents of $p$ in $(a_1,a_2,\dots ,a_{2014})$. Then $Z_p$ is a set of non-negative integers with the following condition: if we remove an arbitrary number from them, then the remaining numbers can be divided into 3 separate groups of 671 elements that have the same sum of all elements. (♣)

We will show that all elements of $Z_p$ are equal. Otherwise, suppose that $z_1,\dots ,z_{2014}$ are not the same, then we can choose a set where the value of $z=z_1+z_2+\cdots +z_{2014}$ is smallest. Note that if we remove $z_i$ then the remaining elements can be divided into three groups of equal sums, hence $z\equiv z_i(\mathrm{mod}3)$ for all $i=1,2,\dots ,2014$. Hence, we investigate three sub-cases.

2.1. Suppose that $z_i\equiv 0(\mathrm{mod}3)$ for all $i=1,2,\dots ,2014$. Then the set $$\left(\frac{z_1}{3},\frac{z_2}{3},\dots ,\frac{z_{2014}}{3}\right)$$ also satisfies condition ♣ with the sum of elements smaller than $z$. This is a contradiction.

2.2. Suppose that $z_i\equiv 1(\mathrm{mod}3)$ for all $i=1,2,\dots ,2014$. Then consider the set $$\left(\frac{z_1+2}{3},\frac{z_2+2}{3},\dots ,\frac{z_{2014}+2}{3}\right),$$ we also have a contradiction.

2.3. Suppose that $z_i\equiv 2(\mathrm{mod}3)$ for all $i=1,2,\dots ,2014$. Consider the set $$\left(\frac{z_1+1}{3},\frac{z_2+1}{3},\dots ,\frac{z_{2014}+2}{3}\right),$$ and we also have a contradiction.

Therefore, we always have $z_1=z_2=\cdots =z_{2014}$ for all $p$. This implies that the absolute values of $a_i$'s are all the same.

Let $0\le k\le 2014$ be the number of negative numbers in $a_i$'s. One can check that all possible values of $k$ are $k\notin \{1,2,2012,2013\}$.

In conclusion, all 2014-tuples satisfying the given condition are: the tuple contains at least 4 zeros, or the tuple contains 2014 numbers of the same nonzero absolute values, and let $0\le k\le 2014$ be the number of negative numbers in the tuple then $k\notin \{1,2,2012,2013\}$.

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