Let's break down the expressions \( B \), \( C \), and \( D \) one by one.

### Calculation of \( B \)
The expression for \( B \) is:
\[
B = 2020 - 2019 + 2018 - 2017 + \ldots + 4 - 3 + 2 + 1
\]
This expression alternates between subtraction and addition of consecutive integers. We can rearrange it to group the pairs:

- Pair \( (2020 - 2019) = 1 \)
- Pair \( (2018 - 2017) = 1 \)
- Continue pairing like this.

The sequence will look like this after pairing:
\[
(2020 - 2019) + (2018 - 2017) + \ldots + (4 - 3) + 2 + 1
\]

The last unpaired terms are 2 and 1.

Now, let's determine the total number of terms:
- The integers from 1 to 2020 form an arithmetic series with the first term as 1 and the last term as 2020. This series contains 2020 terms. Among these:
- Even terms: 2, 4, ..., 2020 (1000 terms)
- Odd terms: 1, 3, ..., 2019 (1000 terms)

Since there are 1000 pairs each contributing 1, plus the last term 1 (because there's an odd number of total terms), we have:
\[
B = 1000 + 2 + 1 = 1001
\]

### Calculation of \( C \)
The expression for \( C \) is:
\[
C = 1 + 3 - 5 + 7 - 9 + \ldots + 2019 + 2021 + 2023
\]
This expression alternates opponents and is less straightforward.

If we carefully examine, we note that:
Pairs are as follows:
- \( (1 + 3) = 4 \)
- \( (7 + 9) = 16 \)
- Further pairs will not have matching subtractions.

The maximum odd number in this range is 2023 which results in:
To count the total number of items:
\[
n = 1012 \ (\text{half from } 1 \text{ to } 2023 \text{ as odd numbers})
\]
Now since the excess is in \( 2019 + 2021 + 2023\).

Thus, calculating as grouped pairs (missing final):
\[
n = ( \text{pairs of } 16) + 2021 + 2023
\]

Calculating the contributions properly will yield, direct evaluations land us to:
\[
C = 1 + 3 - 5 + 7 - 9 + \ldots = 1012 (positive accumulation)
\]

### Calculation of \( D \)
The expression for \( D \) is:
\[
D = 2 - 4 + 6 - 8 + \ldots - 2020 + 2022
\]
This is another alternating sum of even numbers.
The terms are:
- Pairing becomes intuitive, and we continue collapsing.
- Recognizing again pairs yield overall contributions by skipping each \( -4, + 6\),…

Here, notice starts are:
Each section is:
Pairs and each contribute \( -2 \) within every rounding balance.

Calculations:
The series stabilizes through integral components sum.

Now let's total them:
Remember \( 2(k - k+1) = results\):

Summing \( D\):
\[
D = 2 + 0 + 2022 = 1011
\]

### Summary of Results
In conclusion, the calculations yield:
\[
B = 1010\\
C = 1012\\
D = 1011
\]

Thank you.