Solution:
Answer: $89$

Set $m = \frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. Then, in lowest terms, $\frac{N}{10^{k}-1} = \frac{a}{m}$ for some integer $a$. On the other hand, the decimal expansion of $\frac{N}{10^{k}-1}$ simply consists of the decimal expansion of $N$, possibly with some padded zeros, repeating. Since $N$ contains $11235$ as a contiguous substring, the decimal representation of $\frac{a}{m}$ must as well.

Conversely, if $m$ is relatively prime to $10$ and if there exists an $a$ such that the decimal representation of $\frac{a}{m}$ contains the substring $11235$, we claim that $m$ is an attainable value for $\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. To see this, note that since $m$ is relatively prime to $10$, there exists a value of $k$ such that $m$ divides $10^{k}-1$ (for example, $k=\phi(m)$). Letting $ms = 10^{k}-1$ and $N = as$, it follows that $\frac{a}{m} = \frac{as}{ms} = \frac{N}{10^{k}-1}$. Since the decimal expansion of this fraction contains the substring $11235$, it follows that $N$ must also, and therefore $m$ is an attainable value.

We are therefore looking for a fraction $\frac{a}{m}$ which contains the substring $11235$ in its decimal expansion. Since $1,1,2,3$, and $5$ are the first five Fibonacci numbers, it makes sense to look at the value of the infinite series
$$\sum_{i=1}^{\infty} \frac{F_{i}}{10^{i}}$$
A simple generating function argument shows that $\sum_{i=1}^{\infty} F_{i} x^{i} = \frac{x}{1-x-x^{2}}$, so substituting $x=1/10$ leads us to the fraction $10/89$ (which indeed begins $0.11235\ldots$).

How do we know no smaller values of $m$ are possible? Well, if $a'/m'$ contains the substring $11235$ somewhere in its infinitely repeating decimal expansion, then note that there is an $i$ such that the decimal expansion of the fractional part of $10^{i}\left(a'/m'\right)$ begins with $0.11235\ldots$ We can therefore, without loss of generality, assume that the decimal representation of $a'/m'$ begins $0.11235\ldots$ But since the decimal representation of $10/89$ begins $0.11235\ldots$, it follows that
$$\left|\frac{10}{89} - \frac{a'}{m'}\right| \leq 10^{-5}$$
On the other hand, this absolute difference, if non-zero, is at least $\frac{1}{89 m'}$. If $m' < 89$, this is at least $\frac{1}{89^{2}} > 10^{-5}$, and therefore no smaller values of $m'$ are possible.