To solve the problem, we need to find the probability $p$ that for every positive integer $k < 10$, there exists an integer $k' > k$ such that there is at most one number between $k$ and $k'$ in the circle. We will use a recursive approach to determine this probability.

1. Define the Recursive Function:
Let $a_n$ be the number of ways to permute $1, 2, 3, \ldots, n$ cyclically such that the given condition holds. We call a number $a$ "close enough" to $b$ if there is at most one number between $a$ and $b$.

2. Base Cases:
For small values of $n$, we can manually verify the permutations:
- $a_4 = 1$
- $a_5 = 1$

3. Recursive Formula:
We consider two cases to develop the recursive formula for $a_n$:
- Case 1: $n$ and $n-1$ are adjacent. Removing $n$ from the circle leaves a valid permutation of $n-1$ numbers. Thus, there are $2a_{n-1}$ ways to insert $n$ adjacent to $n-1$.
- Case 2: $n$ and $n-1$ have one number between them. Removing both $n$ and the number between them leaves a valid permutation of $n-2$ numbers. There are $2(n-2)a_{n-2}$ ways to insert $n$ and $n-1$ with one number between them.

Combining these cases, we get the recursive relation:
$$a_n = 2a_{n-1} + 2(n-2)a_{n-2}$$

4. Probability Calculation:
Let $p_n = \frac{a_n}{(n-1)!}$. Then:
$$p_n(n-1)! = 2p_{n-1}(n-2)! + 2p_{n-2}(n-2)!$$
Simplifying, we get:
$$p_n = \frac{2}{n-1}(p_{n-1} + p_{n-2})$$

5. **Compute $p_{10}$:**
Using the base cases $p_4 = 1$ and $p_5 = 1$, we compute the subsequent values:
$$p_6 = \frac{2}{5}(p_5 + p_4) = \frac{2}{5}(1 + 1) = \frac{4}{5}$$
$$p_7 = \frac{2}{6}(p_6 + p_5) = \frac{2}{6}\left(\frac{4}{5} + 1\right) = \frac{2}{6} \cdot \frac{9}{5} = \frac{3}{5}$$
$$p_8 = \frac{2}{7}(p_7 + p_6) = \frac{2}{7}\left(\frac{3}{5} + \frac{4}{5}\right) = \frac{2}{7} \cdot \frac{7}{5} = \frac{2}{5}$$
$$p_9 = \frac{2}{8}(p_8 + p_7) = \frac{2}{8}\left(\frac{2}{5} + \frac{3}{5}\right) = \frac{2}{8} \cdot 1 = \frac{1}{4}$$
$$p_{10} = \frac{2}{9}(p_9 + p_8) = \frac{2}{9}\left(\frac{1}{4} + \frac{2}{5}\right) = \frac{2}{9} \cdot \frac{13}{20} = \frac{13}{90}$$

6. Final Calculation:
The probability $p$ is $\frac{13}{90}$. To find $100a + b$ where $p = \frac{a}{b}$ and $a$ and $b$ are relatively prime, we have:
$$100a + b = 100 \cdot 13 + 90 = 1300 + 90 = 1390$$

The final answer is $\boxed{1390}$