Let $p_1, p_2, \ldots, p_{25}$ are primes which don’t exceed 2004. Find the largest integer $T$ such that every positive integer $\leq T$ can be expressed as sums of distinct divisors of $(p_1\cdot p_2 \cdot \ldots \cdot p_{25})^{2004}.$
1. **Define the problem and notation:**
   Let \( p_1, p_2, \ldots, p_{25} \) be primes that do not exceed 2004. We need to find the largest integer \( T \) such that every positive integer \( \leq T \) can be expressed as sums of distinct divisors of \( (p_1 \cdot p_2 \cdot \ldots \cdot p_{25})^{2004} \). Denote \( M = (p_1 p_2 \ldots p_{25})^{2004} \).

2. **Lemma 1:**
   If the sequence \( 1 = a_1 < a_2 < \ldots < a_n \) satisfies the condition \( a_{i+1} \leq 1 + a_1 + \ldots + a_i \) for any \( 1 \leq i \leq n-1 \), then \( a_1 + a_2 + \ldots + a_n \) is the largest number \( T \) such that every positive integer \( \leq T \) can be expressed as sums of distinct terms of the sequence.

   **Proof of Lemma 1:**
   The proof is straightforward. If \( a_{i+1} \leq 1 + a_1 + \ldots + a_i \), then any number up to \( a_1 + a_2 + \ldots + a_n \) can be formed by summing distinct elements from the sequence.

3. **Lemma 2:**
   If the sequence \( 1 = a_1 < a_2 < \ldots < a_n \) satisfies the condition \( a_{i+1} \leq 2a_i \) for any \( 1 \leq i \leq n-1 \), then \( a_1 + a_2 + \ldots + a_n \) is the largest number \( T \) such that every positive integer \( \leq T \) can be expressed as sums of distinct terms of the sequence.

   **Proof of Lemma 2:**
   This follows directly from Lemma 1. If \( a_{i+1} \leq 2a_i \), then \( a_{i+1} \leq 1 + a_1 + \ldots + a_i \) is satisfied because \( 2a_i \leq 1 + a_1 + \ldots + a_i \) for sufficiently large \( i \).

4. **Divisors of \( M \):**
   Let \( 1 = n_1 < n_2 < \ldots < n_N \) be all divisors of \( M \). We need to show that \( n_{i+1} \leq 2n_i \) for any \( 1 \leq i \leq N-1 \).

5. **Case Analysis:**
   - **Case 1:** If \( \min\{p_1, p_2, \ldots, p_{25}\} = 2 \). Suppose \( p_1 = 2 \) and \( n_i = 2^t p_2^{t_2} \ldots p_{25}^{t_{25}} \).
     - If \( t < 2003 \), then \( n_i < 2n_i = 2^{t+1} p_2^{t_2} \ldots p_{25}^{t_{25}} \). Since \( 2n_i \) is a divisor of \( M \), we have \( n_{i+1} \leq 2n_i \).
     - If \( t = 2004 \), then \( \min\{t_2, t_3, \ldots, t_{25}\} \leq 2003 \). Suppose \( t_2 \leq 2003 \). Let \( x \) be the positive integer such that \( 2^x < p_2 < 2^{x+1} \). We have \( n_i < 2^{2004-x} p_2^{t_2+1} \ldots p_{25}^{t_{25}} < 2^{2005} p_2^{t_2} \ldots p_{25}^{t_{25}} \). Since \( 2^{2004-x} p_2^{t_2+1} \ldots p_{25}^{t_{25}} \) is a divisor of \( M \), we have \( n_{i+1} \leq 2n_i \).

6. **Conclusion using Lemma 2:**
   By Lemma 2, we have \( T = n_1 + n_2 + \ldots + n_N = \frac{(p_1^{2005} - 1) \ldots (p_{25}^{2005} - 1)}{(p_1 - 1)(p_2 - 1) \ldots (p_{25} - 1)} \).

7. **Special Case:**
   If \( \min\{p_1, \ldots, p_{25}\} \geq 3 \), then \( M = 1 \).

The final answer is \( \boxed{ \frac{(p_1^{2005} - 1) \ldots (p_{25}^{2005} - 1)}{(p_1 - 1)(p_2 - 1) \ldots (p_{25} - 1)} } \).