Consider the numbers $a_1, a_2, \dots, a_{2014}$, where $a_k$ is the $4k$-digit number whose digits are $k$ groups of the form $2014$, $1 \le k \le 2014$. Since there are only $2013$ possible remainders after the division by $2013$, two different $a$s must give the same remainder in this division, hence there exists $i, j$, with $1 \le j < i \le 2014$, so that $a_i - a_j$ is divisible by $2013$. Since
$$a_i - a_j = \underbrace{20142014\dots2014}_{i-j \text{ times } 2014} \underbrace{00\dots0}_{4j \text{ times}} = \underbrace{20142014\dots2014}_{i-j \text{ times } 2014} \cdot \underbrace{100\dots0}_{4j \text{ times}} = a_{i-j} \cdot 10^{4j}$$
and $10^{4j}$ and $2013$ are relatively prime, $2013$ must divide $a_{i-j}$.