Find the greatest prime that divides $$1^2 - 2^2 + 3^2 - 4^2 +...- 98^2 + 99^2.$$
1. We start with the given series:
   \[
   1^2 - 2^2 + 3^2 - 4^2 + \cdots - 98^2 + 99^2
   \]

2. Notice that we can group the terms in pairs and use the difference of squares formula:
   \[
   a^2 - b^2 = (a - b)(a + b)
   \]
   Grouping the terms, we get:
   \[
   (1^2 - 2^2) + (3^2 - 4^2) + \cdots + (97^2 - 98^2) + 99^2
   \]

3. Applying the difference of squares formula to each pair:
   \[
   (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + \cdots + (97 - 98)(97 + 98) + 99^2
   \]
   Simplifying each term:
   \[
   (-1)(3) + (-1)(7) + \cdots + (-1)(195) + 99^2
   \]
   This simplifies to:
   \[
   -3 - 7 - 11 - \cdots - 195 + 99^2
   \]

4. The series \(-3 - 7 - 11 - \cdots - 195\) is an arithmetic series with the first term \(a = -3\) and common difference \(d = -4\). The number of terms \(n\) can be found by solving:
   \[
   a_n = a + (n-1)d = -195
   \]
   \[
   -3 + (n-1)(-4) = -195
   \]
   \[
   -3 - 4n + 4 = -195
   \]
   \[
   -4n + 1 = -195
   \]
   \[
   -4n = -196
   \]
   \[
   n = 49
   \]

5. The sum of the arithmetic series is given by:
   \[
   S_n = \frac{n}{2} (a + l)
   \]
   where \(l\) is the last term. Here, \(a = -3\), \(l = -195\), and \(n = 49\):
   \[
   S_{49} = \frac{49}{2} (-3 - 195) = \frac{49}{2} (-198) = 49 \times (-99) = -4851
   \]

6. Adding the last term \(99^2\):
   \[
   99^2 = 9801
   \]
   Therefore, the total sum is:
   \[
   -4851 + 9801 = 4950
   \]

7. To find the greatest prime factor of 4950, we perform prime factorization:
   \[
   4950 = 2 \times 3^2 \times 5^2 \times 11
   \]

8. The greatest prime factor is \(11\).

The final answer is \(\boxed{11}\)