What is the smallest possible natural number $n$ for which the equation $x^2 -nx + 2014 = 0$ has integer roots?
1. Let $\zeta$ and $\xi$ be the roots of the quadratic equation $x^2 - nx + 2014 = 0$. By Vieta's formulas, we know:
   \[
   \begin{cases}
   \zeta + \xi = n \\
   \zeta \xi = 2014
   \end{cases}
   \]

2. Since $\zeta$ and $\xi$ are roots of the equation and must be integers, they must be divisors of 2014. We need to find all pairs of integers $(\zeta, \xi)$ such that $\zeta \xi = 2014$.

3. The divisors of 2014 are:
   \[
   2014 = 2 \times 19 \times 53
   \]
   Therefore, the pairs $(\zeta, \xi)$ that satisfy $\zeta \xi = 2014$ are:
   \[
   (\pm 1, \pm 2014), (\pm 2, \pm 1007), (\pm 19, \pm 106), (\pm 38, \pm 53)
   \]

4. We now check each pair to find the smallest possible value of $n = \zeta + \xi$:
   \[
   \begin{aligned}
   &1 + 2014 = 2015 \\
   &-1 - 2014 = -2015 \\
   &1 - 2014 = -2013 \\
   &-1 + 2014 = 2013 \\
   &2 + 1007 = 1009 \\
   &-2 - 1007 = -1009 \\
   &2 - 1007 = -1005 \\
   &-2 + 1007 = 1005 \\
   &19 + 106 = 125 \\
   &-19 - 106 = -125 \\
   &19 - 106 = -87 \\
   &-19 + 106 = 87 \\
   &38 + 53 = 91 \\
   &-38 - 53 = -91 \\
   &38 - 53 = -15 \\
   &-38 + 53 = 15 \\
   \end{aligned}
   \]

5. From the above calculations, the smallest positive value of $n$ is $91$, which occurs when $\zeta = 38$ and $\xi = 53$.

The final answer is $\boxed{91}$