The answer is 48. We observe that after $n$ steps the flea has covered a total distance of $n^{2}$ vertices from the starting point: this is clearly true for $n=0$ and $n=1$, and moreover, if after $n$ steps the flea has covered a distance of $n^{2}$ vertices, then on the next step the distance traveled will be $n^{2}+(2 n+1)=(n+1)^{2}$. Suppose then that the flea touches a vertex twice, the first time when the distance covered is $m^{2}$, the second time when it is $n^{2}$. This happens only if the difference between $n^{2}$ and $m^{2}$ corresponds to an integer number of laps of the 2015-gon, that is, if and only if 2015 divides $n^{2}-m^{2}=(n-m)(n+m)$, and we are looking for the smallest $n$ for which this is possible. Let us take $n, m$ (with $n$ minimal) and write $n-m=a x, n+m=b y$, with $a b=2015$ and $x, y$ positive integers. We now note that $x=y=1$: indeed, the numbers $n^{\prime}=\frac{a+b}{2}, m^{\prime}=\frac{a-b}{2}$ are integers
( $a$ and $b$ are divisors of 2015, so both odd), and satisfy $\left(n^{\prime}+m^{\prime}\right)\left(n^{\prime}-m^{\prime}\right)=a b=2015$, thus they provide a pair $\left(m^{\prime}\right)^{2},\left(n^{\prime}\right)^{2}$ corresponding to a vertex visited twice; now if $x$ or $y$ were greater than 1 we would also have $n^{\prime}=\frac{a+b}{2}<\frac{a x+b y}{2}=n$, which would contradict the minimality of $n$. We therefore want to study the equation $n^{2}-m^{2}=2015=5 \cdot 13 \cdot 31$, and more precisely distribute the prime factors 5, 13, 31 between $n+m$ and $n-m$ so as to minimize $n$. It is then easy to see that the minimum is obtained for $n+m=5 \cdot 13, n-m=31$, which gives $n=48$ and $m=17$.