Problem 1646

1. Initial Observation:
We start by noting that $2021$ can be expressed as the sum of $2021$ squares of $1$:
$$2021 = 1^2 + 1^2 + \cdots + 1^2 \quad (\text{2021 terms})$$
This means $2021$ is an omopeiro number.

2. Reduction Process:
We can reduce the number of terms by substituting four $1^2$ terms with a single $2^2$ term:
$$4 \cdot 1^2 = 2^2$$
This substitution reduces the number of terms by 3 each time we perform it.

3. Constructing New Omopeiro Numbers:
- Start with $2021$:
$$2021 = 1^2 + 1^2 + \cdots + 1^2 \quad (\text{2021 terms})$$
- Substitute four $1^2$ terms with one $2^2$:
$$2021 = 2^2 + 2^2 + \cdots + 2^2 + 1^2 + 1^2 + 1^2 \quad (\text{505 terms of } 2^2 \text{ and 1 term of } 1^2)$$
This gives us $2021 - 3 \times 505 = 2021 - 1515 = 506$ terms.

4. Further Reduction:
- Continue the process:
$$506 = 2^2 + 2^2 + \cdots + 2^2 + 1^2 \quad (\text{126 terms of } 2^2 \text{ and 2 terms of } 1^2)$$
This gives us $506 - 3 \times 126 = 506 - 378 = 128$ terms.
- Continue:
$$128 = 2^2 + 2^2 + \cdots + 2^2 \quad (\text{32 terms of } 2^2)$$
This gives us $128 - 3 \times 32 = 128 - 96 = 32$ terms.
- Continue:
$$32 = 2^2 + 2^2 + \cdots + 2^2 \quad (\text{8 terms of } 2^2)$$
This gives us $32 - 3 \times 8 = 32 - 24 = 8$ terms.
- Continue:
$$8 = 2^2 + 2^2 \quad (\text{2 terms of } 2^2)$$
This gives us $8 - 3 \times 2 = 8 - 6 = 2$ terms.

5. Conclusion:
By repeatedly applying the reduction process, we can generate a sequence of omopeiro numbers. Each step reduces the number of terms by 3, and we can continue this process until we reach a number of terms that is less than or equal to 2.

Since we started with $2021$ and reduced by 3 each time, we can perform this reduction process $\left\lfloor \frac{2021}{3} \right\rfloor = 673$ times. This means we can generate at least $2021 - 3 \times 673 = 2021 - 2019 = 2$ terms.

Therefore, there are at least $2021 - 2 = 2019$ omopeiro numbers.

The final answer is $\boxed{ 2019 }$ omopeiro numbers.