To solve the equation

\[
\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \ldots + \frac{2}{x(x+1)} = \frac{2023}{2025},
\]

we first observe the pattern in the left-hand side. The terms appear to follow a pattern of the form \(\frac{2}{n(n+1)}\) for \(n = 2, 3, \ldots, x\).

Let's simplify the term \(\frac{2}{n(n+1)}\):

\[
\frac{2}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1},
\]

which suggests that the series might telescope. The sum can be rewritten:

\[
S = \sum_{n=2}^{x} \frac{2}{n(n+1)} = \sum_{n=2}^{x} \left( \frac{2}{n} - \frac{2}{n+1} \right).
\]

Now we can expand this into a telescoping series:

\[
S = \left(\frac{2}{2} - \frac{2}{3}\right) + \left(\frac{2}{3} - \frac{2}{4}\right) + \left(\frac{2}{4} - \frac{2}{5}\right) + \ldots + \left(\frac{2}{x} - \frac{2}{x+1}\right).
\]

In this series, all intermediate terms will cancel, leaving us with:

\[
S = 1 - \frac{2}{x+1} = 1 - \frac{2}{x+1}.
\]

Setting this equal to \(\frac{2023}{2025}\):

\[
1 - \frac{2}{x+1} = \frac{2023}{2025}.
\]

Now simplify the left-hand side:

\[
\frac{x+1 - 2}{x+1} = \frac{x - 1}{x + 1} = \frac{2023}{2025}.
\]

Next, we perform cross multiplication:

\[
2025(x - 1) = 2023(x + 1).
\]

Expanding both sides gives:

\[
2025x - 2025 = 2023x + 2023.
\]

Rearranging this leads to:

\[
2025x - 2023x = 2025 + 2023,
\]
\[
2x = 4048.
\]

Dividing both sides by 2:

\[
x = 2024.
\]

Thus, the solution is

\[
\boxed{2024}.
\]