Each of the integers $a_1,a_2,a_3,\dots,a_{2008}$ is at least $1$ and at most $5$. If $a_n < a_{n+1}$, the pair $(a_n, a_{n+1})$ will be called as an increasing pair. If $a_n > a_{n+1}$, the pair $(a_n, a_{n+1})$ will be called as an decreasing pair. If the sequence contains $103$ increasing pairs, at least how many decreasing pairs are there?

$ 
\textbf{(A)}\ 21
\qquad\textbf{(B)}\ 24
\qquad\textbf{(C)}\ 36
\qquad\textbf{(D)}\ 102
\qquad\textbf{(E)}\ \text{None of the above}
$
1. Consider the sequence \(a_1, a_2, a_3, \ldots, a_{2008}\) where each \(a_i\) is an integer between 1 and 5. We are given that there are 103 increasing pairs \((a_n, a_{n+1})\) where \(a_n < a_{n+1}\).

2. Define the sum of the differences for increasing pairs and decreasing pairs:
   \[
   \sum_{a_i > a_{i-1}} (a_i - a_{i-1}) - \sum_{a_j < a_{j-1}} (a_{j-1} - a_j)
   \]
   This sum simplifies to \(a_{2008} - a_1\), which is bounded by the range of the sequence:
   \[
   a_{2008} - a_1 \leq 4
   \]

3. Each term in the first sum (increasing pairs) is at least 1, and there are 103 such terms. Therefore, the first sum is at least 103:
   \[
   \sum_{a_i > a_{i-1}} (a_i - a_{i-1}) \geq 103
   \]

4. Each term in the second sum (decreasing pairs) is at most 4. Let \(k\) be the number of decreasing pairs. Then:
   \[
   \sum_{a_j < a_{j-1}} (a_{j-1} - a_j) \leq 4k
   \]

5. Combining these results, we have:
   \[
   103 - 4k \leq 4
   \]
   Solving for \(k\):
   \[
   103 - 4k \leq 4 \implies 99 \leq 4k \implies k \geq \frac{99}{4} = 24.75
   \]
   Since \(k\) must be an integer, we round up to the nearest whole number:
   \[
   k \geq 25
   \]

6. Therefore, the minimum number of decreasing pairs is 25.

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