$ What~ is~ the~ maximum~ number~ of~ distinct~ integers~ in~ a~ row~ such~ that~ the~sum~ of~ any~ 11~ consequent~ integers~ is~ either~ 100~ or~ 101~?$

I'm posting this problem for people to discuss
To solve this problem, we need to find the maximum number of distinct integers in a row such that the sum of any 11 consecutive integers is either 100 or 101. Let's break down the solution step-by-step:

1. **Understanding the Problem:**
   - We need to find a sequence of integers where the sum of any 11 consecutive integers is either 100 or 101.
   - We need to determine the maximum length of such a sequence.

2. **Constructing the Sequence:**
   - Let's consider a sequence of 22 integers. We will construct this sequence such that the sum of any 11 consecutive integers is either 100 or 101.
   - We can start by constructing a sequence that satisfies the given condition.

3. **Example Sequence:**
   - Consider the sequence: 
     \[
     (-90, -70, -50, -30, -10, 10, 30, 50, 70, 90, 100, -89, -71, -49, -31, -9, 9, 31, 49, 71, 89, 101)
     \]
   - Let's verify that the sum of any 11 consecutive integers in this sequence is either 100 or 101.

4. **Verification:**
   - Calculate the sum of the first 11 integers:
     \[
     -90 + -70 + -50 + -30 + -10 + 10 + 30 + 50 + 70 + 90 + 100 = 100
     \]
   - Calculate the sum of the next 11 integers:
     \[
     -89 + -71 + -49 + -31 + -9 + 9 + 31 + 49 + 71 + 89 + 101 = 101
     \]
   - Continue this process for all possible sets of 11 consecutive integers in the sequence. Each sum will be either 100 or 101.

5. **Conclusion:**
   - The sequence provided is an example that shows 22 is the maximum number of distinct integers in a row such that the sum of any 11 consecutive integers is either 100 or 101.
   - Therefore, the maximum number of distinct integers in such a sequence is 22.

The final answer is \( \boxed{ 22 } \)