An infinite sequence of digits $1$ and $2$ is determined by the following two properties:

i) The sequence is built by writing, in some order, blocks $12$ and blocks $112.$
ii) If each block $12$ is replaced by $1$ and each block $112$ by $2$, the same sequence is again obtained.

In which position is the hundredth digit $1$? What is the thousandth digit of the sequence?

1. **Understanding the Sequence Construction:**
   - The sequence is built using blocks of "12" and "112".
   - If each block "12" is replaced by "1" and each block "112" by "2", the same sequence is obtained.

2. **Defining the Sequences:**
   - Let \( S_1 \) be the original sequence.
   - Let \( S_2 \) be the sequence obtained by replacing "12" with "1" and "112" with "2".

3. **Constructing the Sequences:**
   - Consider the blocks \( A = 12 \) and \( B = 112 \).
   - The sequence \( S_1 \) can be written as a combination of \( A \) and \( B \) in some order.
   - For example, \( S_1 = ABAABABABAB... \).

4. **Counting the Number of 1's in Each Block:**
   - Block \( A \) contains 1's: \( A_1 = 1 \).
   - Block \( B \) contains 1's: \( B_1 = 2 \).

5. **Finding the Position of the 100th Digit 1:**
   - We need to find the position of the 100th digit "1" in the sequence \( S_1 \).
   - Let's consider the sequence \( S_1 \) as a combination of blocks \( A \) and \( B \):
     \[
     S_1 = ABAABABABAB...
     \]
   - The number of 1's in each block:
     \[
     A_1 = 1, \quad B_1 = 2
     \]
   - We need to find the sum of 1's until we reach 100:
     \[
     A_1 + B_1 + A_1 + A_1 + B_1 + A_1 + B_1 + A_1 + B_1 + A_1 + B_1 + A_1 = 1 + 2 + 1 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 17
     \]
   - Repeating this pattern, we find that the 100th digit "1" is in the first position of \( B \) after \( ABAABABABABA \).

6. **Calculating the Position:**
   - The length of the sequence \( ABAABABABABA \) is:
     \[
     |ABAABABABABA| = 12 + 17 + 12 + 12 + 17 + 12 + 17 + 12 + 17 + 12 + 17 + 12 = 169
     \]
   - Therefore, the 100th digit "1" is in the \( 169 + 1 = 170 \)th position.

7. **Finding the 1000th Digit:**
   - We need to find the 1000th digit in the sequence \( S_1 \).
   - The length of the sequence \( S_1 \) is:
     \[
     |S_1| \leq 1000
     \]
   - We can write the sequence as:
     \[
     12 + 17 + 12 + [(12 + 17)k] + m \leq 1000 \quad [m = 0, 12]
     \]
   - Solving for \( k \):
     \[
     29k + m \leq 959 \quad \implies 29 \times 33 = 957 \quad \implies m = 0, k = 33
     \]
   - Therefore, the 1000th digit is the second digit of \( A \), which is \( 2 \).

The final answer is \( \boxed{2} \).