Evaluate the sum \[ 11^2 - 1^1 + 12^2 - 2^2 + 13^2 - 3^2 + \cdots + 20^2 - 10^2. \]
1. We start by rewriting the given sum:
   \[
   11^2 - 1^2 + 12^2 - 2^2 + 13^2 - 3^2 + \cdots + 20^2 - 10^2.
   \]
   This can be expressed as:
   \[
   \sum_{k=1}^{10} (10+k)^2 - k^2.
   \]

2. Next, we simplify each term inside the sum:
   \[
   (10+k)^2 - k^2 = (10+k)(10+k) - k^2 = 100 + 20k + k^2 - k^2 = 100 + 20k.
   \]
   Therefore, the sum becomes:
   \[
   \sum_{k=1}^{10} (100 + 20k).
   \]

3. We can split this sum into two separate sums:
   \[
   \sum_{k=1}^{10} 100 + \sum_{k=1}^{10} 20k.
   \]

4. The first sum is straightforward:
   \[
   \sum_{k=1}^{10} 100 = 100 \times 10 = 1000.
   \]

5. The second sum is the sum of the first 10 positive integers multiplied by 20:
   \[
   \sum_{k=1}^{10} 20k = 20 \sum_{k=1}^{10} k.
   \]
   Using the formula for the sum of the first \( n \) positive integers:
   \[
   \sum_{k=1}^{n} k = \frac{n(n+1)}{2},
   \]
   we get:
   \[
   \sum_{k=1}^{10} k = \frac{10 \cdot 11}{2} = 55.
   \]
   Therefore:
   \[
   20 \sum_{k=1}^{10} k = 20 \times 55 = 1100.
   \]

6. Adding the two sums together:
   \[
   1000 + 1100 = 2100.
   \]

The final answer is \(\boxed{2100}\).