Let the first two terms of the sequence be \( a_1 = 64 \) and \( a_2 = 36 \). Each subsequent term is defined as the average of all preceding terms.

First, we find the third term \( a_3 \):

\[
a_3 = \frac{a_1 + a_2}{2} = \frac{64 + 36}{2} = \frac{100}{2} = 50
\]

Next, we calculate the fourth term \( a_4 \):

\[
a_4 = \frac{a_1 + a_2 + a_3}{3} = \frac{64 + 36 + 50}{3} = \frac{150}{3} = 50
\]

Now we calculate the fifth term \( a_5 \):

\[
a_5 = \frac{a_1 + a_2 + a_3 + a_4}{4} = \frac{64 + 36 + 50 + 50}{4} = \frac{200}{4} = 50
\]

It appears that from \( a_3 \) onwards, the terms stabilize at 50. We can verify this by calculating one more term, \( a_6 \):

\[
a_6 = \frac{a_1 + a_2 + a_3 + a_4 + a_5}{5} = \frac{64 + 36 + 50 + 50 + 50}{5} = \frac{250}{5} = 50
\]

Thus, we observe that starting from \( a_3 \), all subsequent terms are equal to 50:

\[
a_n = 50 \quad \text{for } n \geq 3
\]

Now we calculate the sum of the first 2018 terms. The sum \( S_n \) can be expressed as:

\[
S_{2018} = a_1 + a_2 + a_3 + a_4 + \cdots + a_{2018}
\]

Substituting the values we have:

\[
S_{2018} = a_1 + a_2 + 50 \times (2018 - 2)
\]

Calculating this out:

\[
S_{2018} = 64 + 36 + 50 \times 2016
\]
\[
S_{2018} = 100 + 100800
\]
\[
S_{2018} = 100900
\]

Thus, the sum of the first 2018 terms is

\[
\boxed{100900}
\]