How many positive integer $n$ are there satisfying the inequality $1+\sqrt{n^2-9n+20} > \sqrt{n^2-7n+12}$ ?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{None}$
1. Start with the given inequality:
   \[
   1 + \sqrt{n^2 - 9n + 20} > \sqrt{n^2 - 7n + 12}
   \]

2. Let \( n - 4 = a \). Then, we can rewrite the expressions inside the square roots:
   \[
   n^2 - 9n + 20 = (n-4)(n-5) = a(a-1)
   \]
   \[
   n^2 - 7n + 12 = (n-3)(n-4) = a(a+1)
   \]

3. Substitute these into the inequality:
   \[
   1 + \sqrt{a(a-1)} > \sqrt{a(a+1)}
   \]

4. Square both sides to eliminate the square roots:
   \[
   \left(1 + \sqrt{a(a-1)}\right)^2 > a(a+1)
   \]
   \[
   1 + 2\sqrt{a(a-1)} + a(a-1) > a(a+1)
   \]

5. Simplify the inequality:
   \[
   1 + 2\sqrt{a(a-1)} + a^2 - a > a^2 + a
   \]
   \[
   1 + 2\sqrt{a(a-1)} > 2a
   \]

6. Isolate the square root term:
   \[
   2\sqrt{a(a-1)} > 2a - 1
   \]
   \[
   \sqrt{a(a-1)} > a - \frac{1}{2}
   \]

7. Square both sides again:
   \[
   a(a-1) > \left(a - \frac{1}{2}\right)^2
   \]
   \[
   a^2 - a > a^2 - a + \frac{1}{4}
   \]
   \[
   0 > \frac{1}{4}
   \]

8. This is a contradiction, indicating that our squaring steps might have introduced extraneous solutions. We need to check the values of \( n \) directly for \( n < 5 \) since \( 2a - 1 < 0 \) when \( 2n < 9 \).

9. Test \( n = 1 \):
   \[
   1 + \sqrt{1^2 - 9 \cdot 1 + 20} = 1 + \sqrt{12} > \sqrt{1^2 - 7 \cdot 1 + 12} = \sqrt{6}
   \]
   True.

10. Test \( n = 2 \):
    \[
    1 + \sqrt{2^2 - 9 \cdot 2 + 20} = 1 + \sqrt{6} > \sqrt{2^2 - 7 \cdot 2 + 12} = \sqrt{2}
    \]
    True.

11. Test \( n = 3 \):
    \[
    1 + \sqrt{3^2 - 9 \cdot 3 + 20} = 1 + \sqrt{2} > \sqrt{3^2 - 7 \cdot 3 + 12} = 0
    \]
    True.

12. Test \( n = 4 \):
    \[
    1 + \sqrt{4^2 - 9 \cdot 4 + 20} = 1 + 0 > \sqrt{4^2 - 7 \cdot 4 + 12} = 0
    \]
    True.

13. Therefore, \( n = 1, 2, 3, 4 \) all satisfy the inequality.

The final answer is \(\boxed{4}\)