If $N=1000^2-950^2$, what is the largest prime factor of $N$?

$\textbf{(A) }5\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad\textbf{(E) }29$
1. We start with the given expression \( N = 1000^2 - 950^2 \).

2. Recognize that this is a difference of squares, which can be factored using the identity \( a^2 - b^2 = (a + b)(a - b) \). Here, \( a = 1000 \) and \( b = 950 \).

3. Apply the difference of squares formula:
   \[
   N = 1000^2 - 950^2 = (1000 + 950)(1000 - 950)
   \]

4. Simplify the terms inside the parentheses:
   \[
   N = (1000 + 950)(1000 - 950) = 1950 \cdot 50
   \]

5. Next, we need to find the prime factorization of \( 1950 \) and \( 50 \).

6. Factorize \( 1950 \):
   \[
   1950 = 2 \cdot 3 \cdot 5^2 \cdot 13
   \]

7. Factorize \( 50 \):
   \[
   50 = 2 \cdot 5^2
   \]

8. Combine the prime factorizations:
   \[
   N = 1950 \cdot 50 = (2 \cdot 3 \cdot 5^2 \cdot 13) \cdot (2 \cdot 5^2)
   \]

9. Simplify the combined factorization:
   \[
   N = 2^2 \cdot 3 \cdot 5^4 \cdot 13
   \]

10. Identify the largest prime factor in the factorization \( 2^2 \cdot 3 \cdot 5^4 \cdot 13 \):
    - The prime factors are \( 2, 3, 5, \) and \( 13 \).
    - The largest prime factor is \( 13 \).

The final answer is \(\boxed{13}\).