What is the last two digits of the number $(11^2 + 15^2 + 19^2 +  ...  + 2007^2)^2$?
1. We start by expressing the given sum in a more manageable form. The sequence of numbers is \(11, 15, 19, \ldots, 2007\). This sequence is an arithmetic sequence with the first term \(a = 11\) and common difference \(d = 4\). The general term of the sequence can be written as:
   \[
   a_n = 11 + (n-1) \cdot 4 = 4n + 7
   \]
   Therefore, the sum we need to evaluate is:
   \[
   \left( \sum_{n=1}^{500} (4n + 7)^2 \right)^2
   \]

2. Next, we expand the square inside the sum:
   \[
   (4n + 7)^2 = 16n^2 + 56n + 49
   \]
   Thus, the sum becomes:
   \[
   \sum_{n=1}^{500} (4n + 7)^2 = \sum_{n=1}^{500} (16n^2 + 56n + 49)
   \]

3. We can split the sum into three separate sums:
   \[
   \sum_{n=1}^{500} (16n^2 + 56n + 49) = 16 \sum_{n=1}^{500} n^2 + 56 \sum_{n=1}^{500} n + 49 \sum_{n=1}^{500} 1
   \]

4. We use the formulas for the sum of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:
   \[
   \sum_{n=1}^{500} n = \frac{500(500+1)}{2} = 125250
   \]
   \[
   \sum_{n=1}^{500} n^2 = \frac{500(500+1)(2 \cdot 500 + 1)}{6} = 41791750
   \]
   \[
   \sum_{n=1}^{500} 1 = 500
   \]

5. Substituting these values back into the sum, we get:
   \[
   16 \sum_{n=1}^{500} n^2 + 56 \sum_{n=1}^{500} n + 49 \sum_{n=1}^{500} 1 = 16 \cdot 41791750 + 56 \cdot 125250 + 49 \cdot 500
   \]

6. We now calculate each term modulo 100:
   \[
   16 \cdot 41791750 \equiv 16 \cdot 50 \equiv 800 \equiv 0 \pmod{100}
   \]
   \[
   56 \cdot 125250 \equiv 56 \cdot 50 \equiv 2800 \equiv 0 \pmod{100}
   \]
   \[
   49 \cdot 500 \equiv 49 \cdot 0 \equiv 0 \pmod{100}
   \]

7. Adding these results modulo 100:
   \[
   0 + 0 + 0 \equiv 0 \pmod{100}
   \]

8. Finally, we square the result:
   \[
   (0)^2 \equiv 0 \pmod{100}
   \]

The final answer is \(\boxed{0}\).