Suppose $1/(2n+1) = \dot{a}_1 a_2 \cdots a_n b_1 b_2 \cdots \dot{b}_n$, $a_i + b_i = 9$, $i = 1, 2, \dots, n$, and $2n$ is its period. Since
$$1 + \frac{1}{10^{2n}} + \frac{1}{10^{4n}} + \dots = \frac{1}{1 - \frac{1}{10^{2n}}} = \frac{10^{2n}}{10^{2n} - 1}$$
we have
$$\begin{align*} (1) \quad \frac{1}{2n+1} &= \frac{10^{2n}}{10^{2n}-1} \left( \sum_{i=1}^{n} \frac{a_i}{10^i} + \frac{1}{10^n} \sum_{i=1}^{n} \frac{b_i}{10^i} \right) \\ &= \frac{10^{2n}}{10^{2n}-1} \left( \sum_{i=1}^{n} \frac{a_i}{10^i} + \frac{1}{10^n} \sum_{i=1}^{n} \frac{9-a_i}{10^i} \right) \\ &= \frac{10^{2n}}{10^{2n}-1} \left[ \left(1-\frac{1}{10^n}\right) \sum_{i=1}^{n} \frac{a_i}{10^i} + \frac{9}{10^n} \sum_{i=1}^{n} \frac{1}{10^i} \right] \\ &= \frac{10^{2n}}{10^{2n}-1} \left[ \left(1-\frac{1}{10^n}\right) \sum_{i=1}^{n} \frac{a_i}{10^i} + \frac{9}{10^n} \cdot \frac{1}{10} \cdot \frac{1-\frac{1}{10^n}}{1-\frac{1}{10}} \right] \\ &= \frac{10^{2n}}{10^n+1} \left( \frac{1}{10^n} \sum_{i=1}^{n} \frac{a_i}{10^i} + \frac{1}{10^{2n}} \right) = \frac{\sum_{i=1}^{n} 10^{n-i} a_i + 1}{10^n+1} \end{align*}$$
Therefore,
$2n+1 \mid 10^n + 1.$
By (1) and the knowledge of numbers, we guess the sufficient and necessary condition is
$$(2) \quad 2n+1 \mid 10^n+1 \text{ and } 2n+1 \nmid 10^i+1 \text{ for } i=1,2,\dots,n-1.$$
In fact, since $2n$ is the period of $1/(2n+1)$, we have
$$2n+1 \mid 10^{2n}-1 \text{ and } 2n+1 \nmid 10^i-1 \text{ for } i=1,2,\dots,n-1.$$
So $2n+1 \nmid (10^n+1) + (10^{n+i}-1) = 10^n(10^i+1)$ for $i=1,2,\dots,n-1$. Namely, $2n+1 \nmid 10^i+1$ for $i=1,2,\dots,n-1$. The condition (2) is necessary.

Now we assume that $1/(2n + 1)$ satisfies the condition (2). Let
$$\frac{10^n + 1}{2n + 1} - 1 = \sum_{i=1}^{n} 10^{n-i}a_i,$$
where $0 \le a_i \le 9$, $i = 1, \dots, n$ are integers. Then
$$\frac{1}{2n + 1} = \frac{\sum_{i=1}^{n} 10^{n-i}a_i + 1}{10^n + 1}.$$
Put $b_i = 9 - a_i$. Using (1) we have
$$\frac{1}{2n + 1} = \dot{a}_1 a_2 \cdots a_n b_1 b_2 \cdots \dot{b}_n.$$
For $1 \le i \le n$, because
$$(10^n + 1) + (10^i - 1) = 10^i(10^{n-i} + 1),$$
so $2n + 1 \nmid 10^i - 1$.
For $n < i < 2n$, because
$$(10^n + 1) + (10^i - 1) = 10^n(10^{i-n} + 1),$$
we also have $2n + 1 \nmid 10^i - 1$.
Therefore the period of $1/(2n + 1)$ is $2n$. That is, the condition (2) is sufficient.
By the condition (2), if $1/(2n + 1)$ is a fraction we look for, then $2n + 1$ does not have factors 3 and 5. So the possible values of $1/(2n + 1)$ are 11, 13, 17, 19, 23, .... Since $11, 13 \nmid 10^3 + 1 = 1001$, 11 and 13 do not satisfy the condition. Consider $17 = 2 \times 8 + 1$. We have
$$10^8 + 1 = (17 \times 4 - 2)^4 = 2^4 + 1 = 0 \pmod{17}.$$
For $i = 1, 2, 3, 4$ it is obvious that $17 \nmid 10^i + 1$. For $i = 5, 6, 7$, since $(10^8 + 1) - (10^i + 1) = 10^i(10^{8-i} - 1)$, it is obvious too that $17 \nmid 10^i + 1$. Therefore, $1/17$ is a fraction we need.
Similarly, consider $19 = 2 \times 9 + 1$.
$$10^9 + 1 = (53 \times 19 - 7)^3 + 1 = -342 = 0 \pmod{19}.$$
It is easy to check that $19 \nmid 10^i + 1$ for $1 \le i < 9$. So $1/19$ is another fraction we want to seek.