To evaluate the expression \(-1^2 + 2^2 - 3^2 + 4^2 - \cdots - 19^2 + 20^2\), we need to carefully follow the order of operations and the pattern of signs.

First, let's rewrite the expression more clearly:
\[
-1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - \cdots - 19^2 + 20^2
\]

Notice that the signs alternate between negative and positive, starting with a negative sign for the odd squares and a positive sign for the even squares.

We can separate the sum of the squares of the odd numbers and the squares of the even numbers:
\[
(-1^2 - 3^2 - 5^2 - \cdots - 19^2) + (2^2 + 4^2 + 6^2 + \cdots + 20^2)
\]

Let's calculate the sum of the squares of the odd numbers from 1 to 19:
\[
1^2 + 3^2 + 5^2 + \cdots + 19^2
\]

The sum of the squares of the first \(n\) odd numbers is given by the formula:
\[
\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}
\]

For \(n = 10\) (since there are 10 odd numbers from 1 to 19):
\[
\sum_{k=1}^{10} (2k-1)^2 = \frac{10(20-1)(20+1)}{3} = \frac{10 \cdot 19 \cdot 21}{3} = 1330
\]

Since these terms are all negative in the original expression, we have:
\[
-(1^2 + 3^2 + 5^2 + \cdots + 19^2) = -1330
\]

Next, let's calculate the sum of the squares of the even numbers from 2 to 20:
\[
2^2 + 4^2 + 6^2 + \cdots + 20^2
\]

The sum of the squares of the first \(n\) even numbers is given by the formula:
\[
\sum_{k=1}^n (2k)^2 = 4 \sum_{k=1}^n k^2 = 4 \cdot \frac{n(n+1)(2n+1)}{6}
\]

For \(n = 10\) (since there are 10 even numbers from 2 to 20):
\[
\sum_{k=1}^{10} (2k)^2 = 4 \cdot \frac{10 \cdot 11 \cdot 21}{6} = 4 \cdot 385 = 1540
\]

Since these terms are all positive in the original expression, we have:
\[
2^2 + 4^2 + 6^2 + \cdots + 20^2 = 1540
\]

Combining both results, we get:
\[
-1330 + 1540 = 210
\]

Therefore, the value of the expression \(-1^2 + 2^2 - 3^2 + 4^2 - \cdots - 19^2 + 20^2\) is:
\[
\boxed{210}
\]