a.
If $s(n) = 2$ and $n$ is odd, then $a_n = \frac{1}{2}$. The only solutions with these properties are of the form $n = 10^k + 1$, with $k \in \mathbb{N}^*$.

b.
Let $n$ be a positive integer such that $a_n = \left\{\frac{n}{s(n)}\right\} = \frac{1}{6}$.
Since $\frac{n}{s(n)} - \lfloor \frac{n}{s(n)} \rfloor = \frac{1}{6}$, we infer that $6n - 6 \cdot s(n) \cdot \lfloor \frac{n}{s(n)} \rfloor = s(n)$ (1). From here follows that $6 \mid s(n)$, therefore $3 \mid n$. Consider $n = 3k$ and $s(n) = 6m$, with $m, k$ positive integers. From (1) we deduce that $3k - 6m \cdot \lfloor \frac{k}{2m} \rfloor = m$, hence $3 \mid m$. Consequently $m = 3u$, with $u \in \mathbb{N}^*$ and $s(n) = 18u$, therefore $9 \mid n$.

Consider $n = 9v$, with $v$ a positive integer. From (1) we obtain $3v - 6u \cdot \lfloor \frac{v}{2u} \rfloor = u$, hence $3 \mid u$. Consider $u = 3t$, with $t$ a positive integer. It follows that $m = 9t$ and $s(n) = 54t$, and the minimal sum of the digits of the natural number $n$ is 54.

The smallest positive integer with the sum of its digits 54 is $n = 999999$.

But $a_{999999} = \left\{\frac{999999}{54}\right\} = \frac{1}{2}$, hence $n = 999999$ is not a solution. The next positive integer with the sum of its digits 54 is $n = 1899999$, for which we have $a_{1899999} = \left\{\frac{1899999}{54}\right\} = \{35185 + \frac{1}{6}\} = \frac{1}{6}$, therefore $n_{\min} = 1899999$.