To solve the given expression \((20^2 + 18^2 + 16^2 + \ldots + 4^2 + 2^2) - (19^2 + 17^2 + 15^2 + \ldots + 3^2 + 1^2)\), we need to break it down into manageable parts.

First, let's identify the sequences involved:

1. The first sequence is \(20^2, 18^2, 16^2, \ldots, 4^2, 2^2\).
2. The second sequence is \(19^2, 17^2, 15^2, \ldots, 3^2, 1^2\).

### Step 1: Sum of the first sequence
The first sequence is an arithmetic sequence of even numbers squared:
\[ 20^2, 18^2, 16^2, \ldots, 4^2, 2^2 \]

The general term for this sequence is \((2n)^2\) where \(n\) ranges from 1 to 10 (since \(2 \times 10 = 20\)).

The sum of squares of the first 10 even numbers is:
\[ \sum_{n=1}^{10} (2n)^2 = 4 \sum_{n=1}^{10} n^2 \]

Using the formula for the sum of squares of the first \(n\) natural numbers:
\[ \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \]

For \(k = 10\):
\[ \sum_{n=1}^{10} n^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

Thus:
\[ 4 \sum_{n=1}^{10} n^2 = 4 \cdot 385 = 1540 \]

### Step 2: Sum of the second sequence
The second sequence is an arithmetic sequence of odd numbers squared:
\[ 19^2, 17^2, 15^2, \ldots, 3^2, 1^2 \]

The general term for this sequence is \((2n-1)^2\) where \(n\) ranges from 1 to 10 (since \(2 \times 10 - 1 = 19\)).

The sum of squares of the first 10 odd numbers is:
\[ \sum_{n=1}^{10} (2n-1)^2 \]

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

For \(k = 10\):
\[ \sum_{n=1}^{10} (2n-1)^2 = \frac{10 \cdot 21 \cdot 19}{3} = 1330 \]

### Step 3: Subtract the sums
Now, we subtract the sum of the second sequence from the sum of the first sequence:
\[ 1540 - 1330 = 210 \]

Therefore, the result of the given expression is:
\[ \boxed{210} \]