Let $n\geq 2$ be an integer. For each $k$-tuple of positive integers $a_1, a_2, \ldots, a_k$ such that $a_1+a_2+\cdots +a_k=n$, consider the sums $S_i=1+2+\ldots +a_i$ for $1\leq i\leq k$. Determine, in terms of $n$, the maximum possible value of the product $S_1S_2\cdots S_k$.

[i]Proposed by Misael Pelayo[/i]
1. **Claim 1: There does not exist \(a_i \geq 5\).**

   **Proof:** Suppose there exists \(a_i \geq 5\). Consider replacing \(a_i\) with \(a_j = a_i - 2\) and \(a_l = 2\). We need to show that \(S_i < S_j \cdot S_l\).

   \[
   S_i = \frac{a_i(a_i + 1)}{2}
   \]
   \[
   S_j \cdot S_l = \frac{(a_i - 2)(a_i - 1)}{2} \cdot 3
   \]

   We need to show:
   \[
   \frac{a_i(a_i + 1)}{2} < \frac{3(a_i - 2)(a_i - 1)}{2}
   \]

   Simplifying, this inequality becomes:
   \[
   a_i(a_i + 1) < 3(a_i - 2)(a_i - 1)
   \]

   Expanding both sides:
   \[
   a_i^2 + a_i < 3(a_i^2 - 3a_i + 2)
   \]
   \[
   a_i^2 + a_i < 3a_i^2 - 9a_i + 6
   \]

   Rearranging terms:
   \[
   0 < 2a_i^2 - 10a_i + 6
   \]

   Dividing by 2:
   \[
   0 < a_i^2 - 5a_i + 3
   \]

   The quadratic \(a_i^2 - 5a_i + 3\) is positive for \(a_i \geq 5\). Therefore, \(S_i < S_j \cdot S_l\), contradicting the maximality of the product. Hence, there does not exist \(a_i \geq 5\).

2. **Claim 2: There does not exist \(a_i = 1\).**

   **Proof:** Suppose \(a_i = 1\). Then \(S_i = 1\). Removing this term and adding 1 to another \(a_j\) (since \(n > 1\)), we get a larger product, contradicting the maximality of the product. Hence, there does not exist \(a_i = 1\).

3. **Claim 3: There do not exist two \(a_i, a_j = 2\).**

   **Proof:** Suppose \(a_i = a_j = 2\). Then \(S_i = S_j = 3\). Replacing these with \(a_i = 4\), we get \(S_i = 10\). The product changes from \(3 \cdot 3 \cdot S_1 \cdots S_k\) to \(10 \cdot S_1 \cdots S_k\), which is larger, contradicting the maximality of the product. Hence, there do not exist two \(a_i, a_j = 2\).

4. **Claim 4: There do not exist two \(a_i, a_j = 4\).**

   **Proof:** Suppose \(a_i = a_j = 4\). Replacing these with two \(a_i = 3\) and one \(a_j = 2\), we get \(S_i = S_j = 6\) and \(S_k = 3\). The product changes from \(10 \cdot 10 \cdot S_1 \cdots S_k\) to \(6 \cdot 6 \cdot 3 \cdot S_1 \cdots S_k\), which is larger, contradicting the maximality of the product. Hence, there do not exist two \(a_i, a_j = 4\).

5. **Claim 5: There do not exist \(a_i = 2\) and \(a_j = 4\).**

   **Proof:** Suppose \(a_i = 2\) and \(a_j = 4\). Replacing these with two \(a_i = 3\), we get \(S_i = S_j = 6\). The product changes from \(3 \cdot 10 \cdot S_1 \cdots S_k\) to \(6 \cdot 6 \cdot S_1 \cdots S_k\), which is larger, contradicting the maximality of the product. Hence, there do not exist \(a_i = 2\) and \(a_j = 4\).

From the claims, we obtain that the tuple with the maximum product is:
- \(a) \ 3, 3, \cdots, 3\) if \(n = 3m\)
- \(b) \ 3, 3, \cdots, 3, 4\) if \(n = 3m + 1\)
- \(c) \ 3, 3, \cdots, 3, 2\) if \(n = 3m + 2\)

with products \(6^m\), \(10 \cdot 6^{m-1}\), and \(3 \cdot 6^m\), respectively.

The final answer is \(\boxed{6^m, 10 \cdot 6^{m-1}, 3 \cdot 6^m}\)