For some positive integers $p$, there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C$, $AB=2$, and $CD=AD$.  How many different values of $p<2015$ are possible?

$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$
1. **Define Variables and Setup the Problem:**
   - Let \(AB = 2\), \(BC = h\), \(CD = AD = x\).
   - Let \(H\) be the foot of the perpendicular from \(A\) to \(\overline{CD}\).
   - We need \(h\) and \(x\) to be positive integers, and the perimeter \(p = 2 + h + 2x < 2015\).

2. **Use the Pythagorean Theorem:**
   - Since \( \angle B = 90^\circ \) and \( \angle C = 90^\circ \), \( \triangle AHD \) is a right triangle.
   - \(DH = x - 2\) and \(AH = h\).
   - By the Pythagorean theorem in \(\triangle AHD\):
     \[
     h^2 = x^2 - (x-2)^2
     \]
     \[
     h^2 = x^2 - (x^2 - 4x + 4)
     \]
     \[
     h^2 = 4x - 4
     \]
     \[
     h = 2\sqrt{x-1}
     \]

3. **Determine Integer Conditions:**
   - For \(h\) to be an integer, \(\sqrt{x-1}\) must be an integer.
   - Let \(\sqrt{x-1} = a\), where \(a\) is a positive integer.
   - Then \(x - 1 = a^2\) or \(x = a^2 + 1\).
   - Therefore, \(h = 2a\).

4. **Calculate the Perimeter:**
   - The perimeter \(p\) is given by:
     \[
     p = AB + BC + CD + DA = 2 + h + 2x
     \]
     \[
     p = 2 + 2a + 2(a^2 + 1)
     \]
     \[
     p = 2 + 2a + 2a^2 + 2
     \]
     \[
     p = 2a^2 + 2a + 4
     \]

5. **Find the Range of \(a\):**
   - We need \(p < 2015\):
     \[
     2a^2 + 2a + 4 < 2015
     \]
     \[
     2a^2 + 2a < 2011
     \]
     \[
     a^2 + a < 1005.5
     \]
   - Solve the quadratic inequality \(a^2 + a - 1005.5 < 0\):
     \[
     a^2 + a - 1005.5 = 0
     \]
     - Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
       \[
       a = \frac{-1 \pm \sqrt{1 + 4 \cdot 1005.5}}{2}
       \]
       \[
       a = \frac{-1 \pm \sqrt{4022}}{2}
       \]
       \[
       a \approx \frac{-1 \pm 63.42}{2}
       \]
       - Taking the positive root:
         \[
         a \approx \frac{62.42}{2} \approx 31.21
         \]
   - Since \(a\) must be an integer, the maximum integer value for \(a\) is 31.

6. **Count the Possible Values of \(a\):**
   - The possible values of \(a\) are \(1, 2, \ldots, 31\).
   - There are 31 possible values.

Conclusion:
\[
\boxed{31}
\]