The sequence $ (x_n)$ is defined by $ x_1\equal{}1,x_2\equal{}a$ and $ x_n\equal{}(2n\plus{}1)x_{n\minus{}1}\minus{}(n^2\minus{}1)x_{n\minus{}2}$ $ \forall n \geq 3$, where $ a \in N^*$.For which value of $ a$ does the sequence have the property that $ x_i|x_j$ whenever $ i<j$.
1. **Define the sequence and initial conditions:**
   The sequence \( (x_n) \) is defined by:
   \[
   x_1 = 1, \quad x_2 = a
   \]
   and for \( n \geq 3 \):
   \[
   x_n = (2n+1)x_{n-1} - (n^2-1)x_{n-2}
   \]
   where \( a \in \mathbb{N}^* \).

2. **Determine the condition for \( x_i \mid x_j \) whenever \( i < j \):**
   We need to find the value of \( a \) such that \( x_i \mid x_j \) for all \( i < j \).

3. **Calculate \( x_3 \):**
   Using the recurrence relation for \( n = 3 \):
   \[
   x_3 = (2 \cdot 3 + 1)x_2 - (3^2 - 1)x_{n-2} = 7x_2 - 8
   \]
   For \( x_i \mid x_j \) to hold, \( a \) must divide 8. Thus, \( a \in \{1, 2, 4, 8\} \).

4. **Case 1: \( a = 1 \):**
   \[
   x_1 = 1, \quad x_2 = 1
   \]
   Calculate \( x_4 \):
   \[
   x_4 = (2 \cdot 4 + 1)x_3 - (4^2 - 1)x_2 = 9x_3 - 15x_2
   \]
   Calculate \( x_5 \):
   \[
   x_5 = (2 \cdot 5 + 1)x_4 - (5^2 - 1)x_3 = 11x_4 - 24x_3
   \]
   Check divisibility:
   \[
   x_4 = -24, \quad x_5 = -240
   \]
   For \( x_5 \mid x_6 \):
   \[
   x_6 = 13x_5 - 35x_4
   \]
   \[
   -240 \mid 35 \cdot 24 \quad \text{(not possible)}
   \]
   Hence, \( a = 1 \) does not work.

5. **Case 2: \( a = 2 \):**
   \[
   x_1 = 1, \quad x_2 = 2
   \]
   Calculate \( x_3 \):
   \[
   x_3 = 7 \cdot 2 - 8 = 6
   \]
   Calculate \( x_4 \):
   \[
   x_4 = 9 \cdot 6 - 15 \cdot 2 = 24
   \]
   Calculate \( x_5 \):
   \[
   x_5 = 11 \cdot 24 - 24 \cdot 6 = 120
   \]
   Observe that \( x_n = n! \) for smaller values of \( n \). Prove by induction:
   \[
   x_n = (2n+1)(n-1)! - (n-1)(n+1)(n-2)! = n!
   \]
   Hence, \( x_i \mid x_j \) holds for all \( i < j \).

6. **Case 3: \( a = 4 \):**
   \[
   x_1 = 1, \quad x_2 = 4
   \]
   Calculate \( x_3 \):
   \[
   x_3 = 7 \cdot 4 - 8 = 20
   \]
   Calculate \( x_4 \):
   \[
   x_4 = 9 \cdot 20 - 15 \cdot 4 = 120
   \]
   Calculate \( x_5 \):
   \[
   x_5 = 11 \cdot 120 - 24 \cdot 20 = 720
   \]
   Observe that \( x_n = \frac{(n+2)!}{6} \). Prove by induction:
   \[
   x_n = (2n+1)x_{n-1} - (n^2-1)x_{n-2}
   \]
   Hence, \( x_i \mid x_j \) holds for all \( i < j \).

7. **Case 4: \( a = 8 \):**
   \[
   x_1 = 1, \quad x_2 = 8
   \]
   Calculate \( x_3 \):
   \[
   x_3 = 7 \cdot 8 - 8 = 48
   \]
   Calculate \( x_4 \):
   \[
   x_4 = 9 \cdot 48 - 15 \cdot 8 = 312
   \]
   For \( x_3 \mid x_4 \):
   \[
   48 \mid 15 \cdot 8 = 120 \quad \text{(not possible)}
   \]
   Hence, \( a = 8 \) does not work.

Conclusion:
The values of \( a \) that satisfy the condition are \( a = 2 \) and \( a = 4 \).

The final answer is \( \boxed{ a = 2, 4 } \)