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Denote $f(n)=\displaystyle\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n+1}$ and observe that $f(n)\gt f(n+1)$ for all positive n. Indeed,

$\begin{aligned}\displaystyle f(n)-f(n+1) &= \frac{1}{n+1}-\frac{1}{2n+2}-\frac{1}{2n+3}\\ &= \frac{1}{2n+2}-\frac{1}{2n+3}\\ &\gt 0. \end{aligned}$

This shows that $f$ attains its maximum at $n=1$ . Therefore, suffice it to consider the inequality $\displaystyle f(1)\lt a-2014\frac{1}{7}$ , i.e.,

$\displaystyle \frac{1}{2}+\frac{1}{3}\lt a-2014\frac{1}{7}.$ From here a satisfies $\displaystyle a\gt 2014+\frac{5}{6}+\frac{1}{7}=2014\frac{41}{42}$ . Obviously $a=2015$ is the least integer that satisfies this inequality.