To solve this problem, we need to find the smallest oneic number (a number consisting only of the digit 1) that is divisible by 63. Since 63 can be factored into $63 = 7 \times 9$, the oneic number must be divisible by both 7 and 9.

1. Divisibility by 9:
A number is divisible by 9 if the sum of its digits is divisible by 9. Since a oneic number consists only of the digit 1, the sum of its digits is equal to the number of digits. Therefore, the number of 1's must be a multiple of 9.

2. Divisibility by 7:
To check for divisibility by 7, we need to consider the properties of oneic numbers modulo 7. Let's denote a oneic number with $n$ digits as $N_n$. For example, $N_1 = 1$, $N_2 = 11$, $N_3 = 111$, and so on. We can express $N_n$ as:
$$N_n = \frac{10^n - 1}{9}$$
We need $N_n$ to be divisible by 7. This means:
$$\frac{10^n - 1}{9} \equiv 0 \pmod{7}$$
Simplifying, we get:
$$10^n - 1 \equiv 0 \pmod{63}$$
This implies:
$$10^n \equiv 1 \pmod{63}$$

3. **Finding the smallest $n$**:
We need to find the smallest $n$ such that $10^n \equiv 1 \pmod{63}$. We can use the order of 10 modulo 63 to find this $n$. The order of 10 modulo 63 is the smallest positive integer $d$ such that $10^d \equiv 1 \pmod{63}$.

By testing successive powers of 10 modulo 63, we find:
$$10^1 \equiv 10 \pmod{63}$$
$$10^2 \equiv 100 \equiv 37 \pmod{63}$$
$$10^3 \equiv 370 \equiv 55 \pmod{63}$$
$$10^4 \equiv 550 \equiv 46 \pmod{63}$$
$$10^5 \equiv 460 \equiv 19 \pmod{63}$$
$$10^6 \equiv 190 \equiv 1 \pmod{63}$$
Therefore, the order of 10 modulo 63 is 6. This means the smallest $n$ such that $10^n \equiv 1 \pmod{63}$ is 6.

4. Combining the conditions:
Since $n$ must be a multiple of both 9 (for divisibility by 9) and 6 (for divisibility by 7), the smallest such $n$ is the least common multiple (LCM) of 9 and 6:
$$\text{LCM}(9, 6) = 18$$

Therefore, the smallest oneic number that is divisible by 63 has 18 ones.

The final answer is $\boxed{18}$.