To find the sum of the series \(50 - 49 - 48 - 47 + \ldots + 3 - 1\), we need to carefully analyze the pattern and structure of the series.

First, let's rewrite the series in a more structured way:
\[ 50 - 49 - 48 - 47 + \ldots + 3 - 1 \]

Notice that the series alternates between subtraction and addition, but it starts with a positive number (50) and then subtracts the next three numbers (49, 48, 47), and so on.

Let's break it down step by step:

1. **Group the terms in sets of four:**
\[
(50 - 49 - 48 - 47) + (46 - 45 - 44 - 43) + \ldots + (6 - 5 - 4 - 3) + (2 - 1)
\]

2. **Simplify each group:**
Each group of four terms can be simplified as follows:
\[
(a - (a-1) - (a-2) - (a-3))
\]
Simplifying this:
\[
a - (a-1) - (a-2) - (a-3) = a - a + 1 + 2 + 3 = 1 + 2 + 3 = 6
\]

3. **Count the number of groups:**
The series starts at 50 and ends at 1. To find the number of groups, we note that each group contains 4 terms, except possibly the last group which might have fewer terms.

The number of terms in the series is 50. Dividing by 4 gives:
\[
\frac{50}{4} = 12 \text{ groups with a remainder of 2}
\]
So, there are 12 full groups of 4 terms and 1 additional group with 2 terms (2 and 1).

4. **Sum the full groups:**
Each full group sums to 6, and there are 12 such groups:
\[
12 \times 6 = 72
\]

5. **Sum the remaining terms:**
The remaining terms are 2 and 1:
\[
2 - 1 = 1
\]

6. **Add the sums of the full groups and the remaining terms:**
\[
72 + 1 = 73
\]

Therefore, the sum of the series \(50 - 49 - 48 - 47 + \ldots + 3 - 1\) is \(73\).