Let $N$ be the smallest positive integer $N$ such that $2008N$ is a perfect square and $2007N$ is a perfect cube.  Find the remainder when $N$ is divided by $25$.

$\begin{array}{@{\hspace{-1em}}l@{\hspace{14em}}l@{\hspace{14em}}l} 

\textbf{(A) }0&\textbf{(B) }1&\textbf{(C) }2\\\\ 

\textbf{(D) }3&\textbf{(E) }4&\textbf{(F) }5\\\\

\textbf{(G) }6&\textbf{(H) }7&\textbf{(I) }8\\\\

\textbf{(J) }9&\textbf{(K) }10&\textbf{(L) }11\\\\

\textbf{(M) }12&\textbf{(N) }13&\textbf{(O) }14\\\\

\textbf{(P) }15&\textbf{(Q) }16&\textbf{(R) }17\\\\

\textbf{(S) }18&\textbf{(T) }19&\textbf{(U) }20\\\\

\textbf{(V) }21&\textbf{(W) }22 & \textbf{(X) }23 \end{array}$
To solve the problem, we need to find the smallest positive integer \( N \) such that \( 2008N \) is a perfect square and \( 2007N \) is a perfect cube. We will start by prime factorizing the numbers 2008 and 2007.

1. **Prime Factorization:**
   \[
   2008 = 2^3 \cdot 251
   \]
   \[
   2007 = 3^2 \cdot 223
   \]

2. **Conditions for \( N \):**
   - For \( 2008N \) to be a perfect square, all the exponents in its prime factorization must be even.
   - For \( 2007N \) to be a perfect cube, all the exponents in its prime factorization must be multiples of 3.

3. **Prime Factorization of \( N \):**
   Let \( N \) be expressed as:
   \[
   N = 2^a \cdot 3^b \cdot 223^c \cdot 251^d
   \]

4. **Perfect Square Condition:**
   \[
   2008N = 2^{3+a} \cdot 3^b \cdot 223^c \cdot 251^{1+d}
   \]
   For \( 2008N \) to be a perfect square:
   \[
   3 + a \text{ must be even} \implies a \text{ must be odd}
   \]
   \[
   b \text{ must be even}
   \]
   \[
   c \text{ must be even}
   \]
   \[
   1 + d \text{ must be even} \implies d \text{ must be odd}
   \]

5. **Perfect Cube Condition:**
   \[
   2007N = 2^a \cdot 3^{2+b} \cdot 223^{1+c} \cdot 251^d
   \]
   For \( 2007N \) to be a perfect cube:
   \[
   a \text{ must be a multiple of 3}
   \]
   \[
   2 + b \text{ must be a multiple of 3} \implies b \equiv 1 \pmod{3}
   \]
   \[
   1 + c \text{ must be a multiple of 3} \implies c \equiv 2 \pmod{3}
   \]
   \[
   d \text{ must be a multiple of 3}
   \]

6. **Finding the Smallest \( N \):**
   - \( a \) must be the smallest odd multiple of 3, so \( a = 3 \).
   - \( b \) must be the smallest even number such that \( b \equiv 1 \pmod{3} \), so \( b = 4 \).
   - \( c \) must be the smallest even number such that \( c \equiv 2 \pmod{3} \), so \( c = 2 \).
   - \( d \) must be the smallest odd multiple of 3, so \( d = 3 \).

   Therefore:
   \[
   N = 2^3 \cdot 3^4 \cdot 223^2 \cdot 251^3
   \]

7. **Finding the Remainder of \( N \) modulo 25:**
   \[
   2^3 = 8
   \]
   \[
   3^4 = 81 \equiv 6 \pmod{25}
   \]
   \[
   223 \equiv -2 \pmod{25} \implies 223^2 \equiv (-2)^2 = 4 \pmod{25}
   \]
   \[
   251 \equiv 1 \pmod{25} \implies 251^3 \equiv 1^3 = 1 \pmod{25}
   \]

   Combining these:
   \[
   N \equiv 8 \cdot 6 \cdot 4 \cdot 1 \pmod{25}
   \]
   \[
   N \equiv 8 \cdot 6 = 48 \equiv 23 \pmod{25}
   \]
   \[
   23 \cdot 4 = 92 \equiv 17 \pmod{25}
   \]

The final answer is \( \boxed{17} \).