Determine the number of all numbers which are represented as $x^2+y^2$ with $x, y \in \{1, 2, 3, \ldots, 1000\}$ and which are divisible by 121.
1. **Identify the quadratic residues modulo 11:**
   The quadratic residues modulo 11 are the possible remainders when squares of integers are divided by 11. These residues are:
   \[
   \{0, 1, 4, 9, 5, 3\}
   \]
   This can be verified by squaring each integer from 0 to 10 and taking the result modulo 11.

2. **Determine when the sum of two quadratic residues is divisible by 11:**
   We need to find pairs \((x^2, y^2)\) such that:
   \[
   x^2 + y^2 \equiv 0 \pmod{11}
   \]
   By examining the set of quadratic residues, we see that no two non-zero elements from the set sum to 0 modulo 11. Therefore, the only way for \(x^2 + y^2\) to be 0 modulo 11 is if both \(x^2\) and \(y^2\) are 0 modulo 11. This implies:
   \[
   11 \mid x \quad \text{and} \quad 11 \mid y
   \]

3. **Count the number of valid pairs \((x, y)\):**
   Since \(x\) and \(y\) must be multiples of 11, we can write:
   \[
   x = 11a \quad \text{and} \quad y = 11b
   \]
   where \(a, b \in \{1, 2, 3, \ldots, 90\}\). This is because \(x\) and \(y\) range from 1 to 1000, and there are 90 multiples of 11 within this range.

4. **Calculate the number of such pairs:**
   Each of \(a\) and \(b\) can independently take any of the 90 values. Therefore, the total number of pairs \((a, b)\) is:
   \[
   90 \times 90 = 8100
   \]

5. **Conclusion:**
   The number of ordered pairs \((x, y)\) such that \(x^2 + y^2\) is divisible by 121 is 8100.

The final answer is \(\boxed{8100}\)