To solve the series \( \frac{1}{1000} + \frac{13}{1000} + \frac{25}{1000} + \frac{37}{1000} + \frac{49}{1000} + \ldots + \frac{109}{1000} \), we first need to identify the sequence in the numerators.

The numerators form the sequence:
1, 13, 25, 37, 49, ..., 109.

We can see that the difference between consecutive terms is constant. To find the difference:

- \( 13 - 1 = 12 \)
- \( 25 - 13 = 12 \)
- \( 37 - 25 = 12 \)
- \( 49 - 37 = 12 \)

Thus, the sequence is an arithmetic progression with a first term \( a = 1 \) and a common difference \( d = 12 \).

To find the number of terms \( n \) in the sequence, we can use the \( n \)-th term formula for an arithmetic series:

\[
a_n = a + (n-1)d
\]
Setting \( a_n = 109 \):

\[
109 = 1 + (n-1) \cdot 12
\]

Subtracting 1 from both sides:

\[
108 = (n-1) \cdot 12
\]

Dividing both sides by 12:

\[
n - 1 = 9
\]
\[
n = 10
\]

So, there are 10 terms in the series.

Next, we can find the sum of the series using the sum formula for an arithmetic progression:

\[
S_n = \frac{n}{2} (a + a_n)
\]

Substituting \( n = 10 \), \( a = 1 \), and \( a_n = 109 \):

\[
S_{10} = \frac{10}{2} (1 + 109) = 5 \times 110 = 550
\]

This 550 represents the sum of the numerators. Since each term in the series involves division by 1000, we multiply the total sum by \( \frac{1}{1000} \):

\[
\frac{S_{10}}{1000} = \frac{550}{1000} = 0.55
\]

Thus, the sum of the series is:

\[
\boxed{0.55}
\]