Let n be the natural number ($n\ge 2$). All natural numbers from $1$  up to $n$ ,inclusive, are written on the board   in some order: $a_1$, $a_2$ , $...$ , $a_n$. Determine all natural numbers $n$ ($n\ge 2$), for which the product
$$P = (1 + a_1) \cdot (2 + a_2) \cdot ... \cdot (n + a_n)$$
 is an even number, whatever the arrangement of the numbers written on the board.
1. **Case 1: \( n \) is even**

   If \( n \) is even, we can construct a specific arrangement of the numbers \( a_1, a_2, \ldots, a_n \) such that the product \( P \) is odd. Consider the arrangement where \( a_i = i \) for \( i = 1, 2, \ldots, n \). Then, the product \( P \) becomes:
   \[
   P = (1 + a_1) \cdot (2 + a_2) \cdot \ldots \cdot (n + a_n) = (1 + 1) \cdot (2 + 2) \cdot \ldots \cdot (n + n)
   \]
   Simplifying, we get:
   \[
   P = 2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2n
   \]
   Since each term in the product is even, the entire product \( P \) is even. Therefore, for even \( n \), it is possible to arrange the numbers such that \( P \) is even.

2. **Case 2: \( n \) is odd**

   If \( n \) is odd, we need to show that the product \( P \) is always even, regardless of the arrangement of the numbers \( a_1, a_2, \ldots, a_n \). Let \( b_i = i + a_i \) for \( i = 1, 2, \ldots, n \). Observe that:
   \[
   \sum_{i=1}^n b_i = \sum_{i=1}^n (i + a_i) = \sum_{i=1}^n i + \sum_{i=1}^n a_i
   \]
   Since \( a_i \) is a permutation of \( \{1, 2, \ldots, n\} \), we have:
   \[
   \sum_{i=1}^n a_i = \sum_{i=1}^n i = \frac{n(n+1)}{2}
   \]
   Therefore:
   \[
   \sum_{i=1}^n b_i = \sum_{i=1}^n i + \sum_{i=1}^n a_i = \frac{n(n+1)}{2} + \frac{n(n+1)}{2} = n(n+1)
   \]
   Since \( n \) is odd, \( n+1 \) is even, and thus \( n(n+1) \) is even. This implies that the sum \( \sum_{i=1}^n b_i \) is even.

   If all \( b_i \) were odd, then the sum \( \sum_{i=1}^n b_i \) would be odd (since the sum of an odd number of odd numbers is odd), which contradicts the fact that \( \sum_{i=1}^n b_i \) is even. Therefore, at least one \( b_i \) must be even, making the product \( P \) even.

Conclusion:
For \( n \) even, it is possible to arrange the numbers such that \( P \) is even. For \( n \) odd, the product \( P \) is always even, regardless of the arrangement.

The final answer is \( \boxed{ n } \) is odd.