Suppose that $P(x)$ is a monic cubic polynomial with integer roots, and suppose that $\frac{P(a)}{a}$ is an integer for exactly $6$ integer values of $a$. Suppose furthermore that exactly one of the distinct numbers $\frac{P(1) + P(-1)}{2}$ and $\frac{P(1) - P(-1)}{2}$ is a perfect square. Given that $P(0) > 0$, find the second-smallest possible value of $P(0).$

[i]Proposed by Andrew Wu[/i]
1. Let \( P(x) = x^3 + dx^2 + ex + f \), where \( d \), \( e \), and \( f \) are integers. Since \( P(x) \) is a monic cubic polynomial with integer roots, it can be written as \( P(x) = (x - r_1)(x - r_2)(x - r_3) \) for some integers \( r_1, r_2, r_3 \).

2. We are given that \( \frac{P(a)}{a} \) is an integer for exactly 6 integer values of \( a \). This implies that \( P(a) = a(a^2 + da + e) + f \) must be divisible by \( a \) for these 6 values. Therefore, \( f \) must have exactly 6 divisors (considering both positive and negative values). The only numbers with exactly 6 divisors are the squares of primes, i.e., \( f = p^2 \) for some prime \( p \).

3. We are also given that exactly one of the numbers \( \frac{P(1) + P(-1)}{2} \) and \( \frac{P(1) - P(-1)}{2} \) is a perfect square. We calculate:
   \[
   P(1) = 1 + d + e + f
   \]
   \[
   P(-1) = -1 + d - e + f
   \]
   \[
   \frac{P(1) + P(-1)}{2} = \frac{(1 + d + e + f) + (-1 + d - e + f)}{2} = d + f
   \]
   \[
   \frac{P(1) - P(-1)}{2} = \frac{(1 + d + e + f) - (-1 + d - e + f)}{2} = 1 + e
   \]

4. Since exactly one of \( d + f \) and \( 1 + e \) is a perfect square, we need to consider the cases where one is a perfect square and the other is not.

5. We also know that \( P(0) = f \) and \( P(0) > 0 \). Therefore, \( f = p^2 \) for some prime \( p \).

6. We need to find the second-smallest possible value of \( P(0) \). We start by testing small primes:
   - For \( p = 2 \):
     \[
     f = 2^2 = 4
     \]
     \[
     d + f = d + 4
     \]
     \[
     1 + e = 1 + e
     \]
     We need to check if either \( d + 4 \) or \( 1 + e \) is a perfect square. If \( d + 4 \) is a perfect square, let \( d + 4 = k^2 \) for some integer \( k \). Then \( d = k^2 - 4 \). If \( 1 + e \) is a perfect square, let \( 1 + e = m^2 \) for some integer \( m \). Then \( e = m^2 - 1 \).

   - For \( p = 3 \):
     \[
     f = 3^2 = 9
     \]
     \[
     d + f = d + 9
     \]
     \[
     1 + e = 1 + e
     \]
     We need to check if either \( d + 9 \) or \( 1 + e \) is a perfect square. If \( d + 9 \) is a perfect square, let \( d + 9 = k^2 \) for some integer \( k \). Then \( d = k^2 - 9 \). If \( 1 + e \) is a perfect square, let \( 1 + e = m^2 \) for some integer \( m \). Then \( e = m^2 - 1 \).

   - For \( p = 5 \):
     \[
     f = 5^2 = 25
     \]
     \[
     d + f = d + 25
     \]
     \[
     1 + e = 1 + e
     \]
     We need to check if either \( d + 25 \) or \( 1 + e \) is a perfect square. If \( d + 25 \) is a perfect square, let \( d + 25 = k^2 \) for some integer \( k \). Then \( d = k^2 - 25 \). If \( 1 + e \) is a perfect square, let \( 1 + e = m^2 \) for some integer \( m \). Then \( e = m^2 - 1 \).

7. After testing these values, we find that the second-smallest possible value of \( P(0) \) is when \( p = 5 \), giving \( P(0) = 25 \).

The final answer is \( \boxed{25} \)