What is the smallest positive integer $k$ such that the number $\textstyle\binom{2k}k$ ends in two zeros?

$\textbf{(A) }3\hspace{14em}\textbf{(B) }4\hspace{14em}\textbf{(C) }5$

$\textbf{(D) }6\hspace{14em}\textbf{(E) }7\hspace{14em}\textbf{(F) }8$

$\textbf{(G) }9\hspace{14em}\textbf{(H) }10\hspace{13.3em}\textbf{(I) }11$

$\textbf{(J) }12\hspace{13.8em}\textbf{(K) }13\hspace{13.3em}\textbf{(L) }14$

$\textbf{(M) }2007$
To determine the smallest positive integer \( k \) such that the number \( \binom{2k}{k} \) ends in two zeros, we need to ensure that \( \binom{2k}{k} \) is divisible by \( 100 \). This means that \( \binom{2k}{k} \) must have at least two factors of 2 and two factors of 5 in its prime factorization.

Recall that the binomial coefficient is given by:
\[
\binom{2k}{k} = \frac{(2k)!}{(k!)^2}
\]

To find the number of trailing zeros in \( \binom{2k}{k} \), we need to count the number of factors of 2 and 5 in \( \frac{(2k)!}{(k!)^2} \).

1. **Count the factors of 2 in \( \binom{2k}{k} \):**
   The number of factors of 2 in \( n! \) is given by:
   \[
   \sum_{i=1}^{\infty} \left\lfloor \frac{n}{2^i} \right\rfloor
   \]
   Therefore, the number of factors of 2 in \( (2k)! \) is:
   \[
   \sum_{i=1}^{\infty} \left\lfloor \frac{2k}{2^i} \right\rfloor
   \]
   The number of factors of 2 in \( (k!)^2 \) is:
   \[
   2 \sum_{i=1}^{\infty} \left\lfloor \frac{k}{2^i} \right\rfloor
   \]
   Thus, the number of factors of 2 in \( \binom{2k}{k} \) is:
   \[
   \sum_{i=1}^{\infty} \left\lfloor \frac{2k}{2^i} \right\rfloor - 2 \sum_{i=1}^{\infty} \left\lfloor \frac{k}{2^i} \right\rfloor
   \]

2. **Count the factors of 5 in \( \binom{2k}{k} \):**
   Similarly, the number of factors of 5 in \( (2k)! \) is:
   \[
   \sum_{i=1}^{\infty} \left\lfloor \frac{2k}{5^i} \right\rfloor
   \]
   The number of factors of 5 in \( (k!)^2 \) is:
   \[
   2 \sum_{i=1}^{\infty} \left\lfloor \frac{k}{5^i} \right\rfloor
   \]
   Thus, the number of factors of 5 in \( \binom{2k}{k} \) is:
   \[
   \sum_{i=1}^{\infty} \left\lfloor \frac{2k}{5^i} \right\rfloor - 2 \sum_{i=1}^{\infty} \left\lfloor \frac{k}{5^i} \right\rfloor
   \]

3. **Find the smallest \( k \) such that \( \binom{2k}{k} \) has at least 2 factors of 2 and 2 factors of 5:**
   We need to find the smallest \( k \) such that both the number of factors of 2 and the number of factors of 5 in \( \binom{2k}{k} \) are at least 2.

   After calculating for different values of \( k \), we find that:
   - For \( k = 13 \):
     \[
     \sum_{i=1}^{\infty} \left\lfloor \frac{26}{2^i} \right\rfloor - 2 \sum_{i=1}^{\infty} \left\lfloor \frac{13}{2^i} \right\rfloor = 22 - 20 = 2
     \]
     \[
     \sum_{i=1}^{\infty} \left\lfloor \frac{26}{5^i} \right\rfloor - 2 \sum_{i=1}^{\infty} \left\lfloor \frac{13}{5^i} \right\rfloor = 5 - 3 = 2
     \]
     Both conditions are satisfied.

Therefore, the smallest positive integer \( k \) such that \( \binom{2k}{k} \) ends in two zeros is \( k = 13 \).

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