For each positive integer $n$, let $S_n = \sum_{k=1}^nk^3$, and let $d(n)$ be the number of positive divisors of $n$.  For how many positive integers $m$, where $m\leq 25$, is there a solution $n$ to the equation $d(S_n) = m$?
To solve the problem, we need to find the number of positive integers \( m \) such that \( m \leq 25 \) and there exists a positive integer \( n \) for which \( d(S_n) = m \). Here, \( S_n = \sum_{k=1}^n k^3 \) and \( d(n) \) is the number of positive divisors of \( n \).

First, we recall the formula for the sum of cubes:
\[ S_n = \left( \frac{n(n+1)}{2} \right)^2 \]

Next, we need to find the number of divisors of \( S_n \):
\[ d(S_n) = d\left( \left( \frac{n(n+1)}{2} \right)^2 \right) \]

To find \( d(S_n) \), we need to factorize \( \frac{n(n+1)}{2} \). Let:
\[ \frac{n(n+1)}{2} = \prod_{i=1}^r p_i^{a_i} \]
where \( p_i \) are prime numbers and \( a_i \geq 1 \).

Then:
\[ S_n = \left( \frac{n(n+1)}{2} \right)^2 = \left( \prod_{i=1}^r p_i^{a_i} \right)^2 = \prod_{i=1}^r p_i^{2a_i} \]

The number of divisors of \( S_n \) is given by:
\[ d(S_n) = \prod_{i=1}^r (2a_i + 1) \]

We need to find \( m \) such that:
\[ \prod_{i=1}^r (2a_i + 1) \leq 25 \]

Let's consider different values of \( r \) and find the corresponding \( m \):

1. **Case \( r = 1 \):**
   \[ d(S_n) = 2a_1 + 1 \]
   We need \( 2a_1 + 1 \leq 25 \):
   \[ 2a_1 \leq 24 \]
   \[ a_1 \leq 12 \]
   Possible values of \( m \) are \( 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 \).

2. **Case \( r = 2 \):**
   \[ d(S_n) = (2a_1 + 1)(2a_2 + 1) \]
   We need \( (2a_1 + 1)(2a_2 + 1) \leq 25 \):
   - If \( a_1 = 1 \), \( 2a_1 + 1 = 3 \):
     \[ 3(2a_2 + 1) \leq 25 \]
     \[ 2a_2 + 1 \leq \frac{25}{3} \approx 8.33 \]
     \[ 2a_2 \leq 7 \]
     \[ a_2 \leq 3 \]
     Possible values of \( m \) are \( 3 \times 3 = 9 \), \( 3 \times 5 = 15 \), \( 3 \times 7 = 21 \).

   - If \( a_1 = 2 \), \( 2a_1 + 1 = 5 \):
     \[ 5(2a_2 + 1) \leq 25 \]
     \[ 2a_2 + 1 \leq 5 \]
     \[ 2a_2 \leq 4 \]
     \[ a_2 \leq 2 \]
     Possible values of \( m \) are \( 5 \times 3 = 15 \), \( 5 \times 5 = 25 \).

3. **Case \( r = 3 \):**
   \[ d(S_n) = (2a_1 + 1)(2a_2 + 1)(2a_3 + 1) \]
   We need \( (2a_1 + 1)(2a_2 + 1)(2a_3 + 1) \leq 25 \):
   - If \( a_1 = 1 \), \( 2a_1 + 1 = 3 \):
     \[ 3(2a_2 + 1)(2a_3 + 1) \leq 25 \]
     - If \( a_2 = 1 \), \( 2a_2 + 1 = 3 \):
       \[ 3 \times 3(2a_3 + 1) \leq 25 \]
       \[ 9(2a_3 + 1) \leq 25 \]
       \[ 2a_3 + 1 \leq \frac{25}{9} \approx 2.78 \]
       \[ 2a_3 \leq 1 \]
       \[ a_3 \leq 0 \]
       No valid \( a_3 \).

Thus, the possible values of \( m \) are \( 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 \).

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