For each positive integer $k$ greater than $1$, find the largest real number $t$ such that the following hold:
Given $n$ distinct points $a^{(1)}=(a^{(1)}_1,\ldots, a^{(1)}_k)$, $\ldots$, $a^{(n)}=(a^{(n)}_1,\ldots, a^{(n)}_k)$ in $\mathbb{R}^k$, we define the score of the tuple $a^{(i)}$ as
\[\prod_{j=1}^{k}\#\{1\leq i'\leq n\textup{ such that }\pi_j(a^{(i')})=\pi_j(a^{(i)})\}\]
where $\#S$ is the number of elements in set $S$, and $\pi_j$ is the projection $\mathbb{R}^k\to \mathbb{R}^{k-1}$ omitting the $j$-th coordinate. Then the $t$-th power mean of the scores of all $a^{(i)}$'s is at most $n$.

Note: The $t$-th power mean of positive real numbers $x_1,\ldots,x_n$ is defined as 
\[\left(\frac{x_1^t+\cdots+x_n^t}{n}\right)^{1/t}\]
when $t\neq 0$, and it is $\sqrt[n]{x_1\cdots x_n}$ when $t=0$.

[i]Proposed by Cheng-Ying Chang and usjl[/i]
To solve this problem, we need to find the largest real number \( t \) such that the \( t \)-th power mean of the scores of all \( a^{(i)} \)'s is at most \( n \). Let's break down the problem step by step.

1. **Define the Score:**
   Given \( n \) distinct points \( a^{(1)}, a^{(2)}, \ldots, a^{(n)} \) in \( \mathbb{R}^k \), the score of the tuple \( a^{(i)} \) is defined as:
   \[
   \prod_{j=1}^{k} \#\{1 \leq i' \leq n \text{ such that } \pi_j(a^{(i')}) = \pi_j(a^{(i)})\}
   \]
   Here, \( \pi_j \) is the projection from \( \mathbb{R}^k \) to \( \mathbb{R}^{k-1} \) omitting the \( j \)-th coordinate.

2. **Understanding the Projection:**
   The projection \( \pi_j(a^{(i)}) \) omits the \( j \)-th coordinate of \( a^{(i)} \). For each \( j \), we count how many points \( a^{(i')} \) have the same projection as \( a^{(i)} \).

3. **Score Calculation:**
   The score of \( a^{(i)} \) is the product of these counts over all \( j \) from 1 to \( k \).

4. **Power Mean:**
   The \( t \)-th power mean of the scores is given by:
   \[
   \left( \frac{x_1^t + x_2^t + \cdots + x_n^t}{n} \right)^{1/t}
   \]
   where \( x_i \) is the score of \( a^{(i)} \).

5. **Bounding the Power Mean:**
   We need to show that this \( t \)-th power mean is at most \( n \). This translates to:
   \[
   \left( \frac{x_1^t + x_2^t + \cdots + x_n^t}{n} \right)^{1/t} \leq n
   \]
   Simplifying, we get:
   \[
   \frac{x_1^t + x_2^t + \cdots + x_n^t}{n} \leq n^t
   \]
   \[
   x_1^t + x_2^t + \cdots + x_n^t \leq n^{t+1}
   \]

6. **Using the Joints Theorem:**
   According to the Joints Theorem, the \( 1/(k-1) \)-th power mean of the scores is at most \( k!n \). However, in our case, the lines are parallel to the axes, which simplifies the problem.

7. **Parameter Counting:**
   By considering polynomials of degree at most \( n \) in each variable, we can replace the \( k! \) factor with a simpler bound. This leads to the conclusion that the \( 1/(k-1) \)-th power mean of the scores is at most \( n \).

8. **Conclusion:**
   Therefore, the largest \( t \) such that the \( t \)-th power mean of the scores is at most \( n \) is \( t = \frac{1}{k-1} \).

The final answer is \( \boxed{ t = \frac{1}{k-1} } \).