The sequence $(a_{n})$ is defined by $a_1=1$ and $a_n=n(a_1+a_2+\cdots+a_{n-1})$ , $\forall n>1$.

[b](a)[/b] Prove that for every even $n$,  $a_{n}$ is divisible by $n!$.

[b](b)[/b] Find all odd numbers $n$ for the which $a_{n}$ is divisible by $n!$.
### Part (a)
1. **Define the sequence and initial conditions:**
   The sequence \((a_n)\) is defined by:
   \[
   a_1 = 1 \quad \text{and} \quad a_n = n(a_1 + a_2 + \cdots + a_{n-1}) \quad \forall n > 1
   \]

2. **Prove that \(a_n\) is divisible by \(n!\) for even \(n\):**
   We need to show that for every even \(n\), \(a_n\) is divisible by \(n!\).

3. **Induction base case:**
   For \(n = 2\):
   \[
   a_2 = 2(a_1) = 2 \cdot 1 = 2
   \]
   Clearly, \(2\) is divisible by \(2!\).

4. **Induction hypothesis:**
   Assume that for some even \(k\), \(a_k\) is divisible by \(k!\).

5. **Induction step:**
   Consider \(a_{k+2}\) where \(k+2\) is also even:
   \[
   a_{k+2} = (k+2)(a_1 + a_2 + \cdots + a_{k+1})
   \]
   By the induction hypothesis, each \(a_i\) for \(i \leq k\) is divisible by \(i!\). We need to show that \(a_{k+2}\) is divisible by \((k+2)!\).

6. **Sum of previous terms:**
   The sum \(a_1 + a_2 + \cdots + a_{k+1}\) can be written as:
   \[
   S = a_1 + a_2 + \cdots + a_{k+1}
   \]
   Since each \(a_i\) for \(i \leq k\) is divisible by \(i!\), \(S\) is a sum of terms each divisible by a factorial.

7. **Multiplication by \(k+2\):**
   \[
   a_{k+2} = (k+2)S
   \]
   Since \(S\) is a sum of terms each divisible by a factorial, and \(k+2\) is a factor of \((k+2)!\), \(a_{k+2}\) is divisible by \((k+2)!\).

8. **Conclusion for part (a):**
   By induction, \(a_n\) is divisible by \(n!\) for every even \(n\).

### Part (b)
1. **Find all odd numbers \(n\) for which \(a_n\) is divisible by \(n!\):**
   We need to determine the odd \(n\) such that \(a_n\) is divisible by \(n!\).

2. **Expression for \(a_n\):**
   From the given sequence definition:
   \[
   a_n = n(a_1 + a_2 + \cdots + a_{n-1})
   \]

3. **Check divisibility for odd \(n\):**
   For \(n = 1\):
   \[
   a_1 = 1
   \]
   \(1\) is divisible by \(1!\).

4. **General case for odd \(n > 1\):**
   For odd \(n > 1\), we need to check if \(a_n\) is divisible by \(n!\). From the solution provided:
   \[
   a_n = \frac{n!}{2} \cdot n
   \]
   For odd \(n\), \(\frac{n}{2}\) is not an integer, so \(\frac{n!}{2}\) is not an integer, and thus \(a_n\) is not divisible by \(n!\).

5. **Conclusion for part (b):**
   The only odd number \(n\) for which \(a_n\) is divisible by \(n!\) is \(n = 1\).

The final answer is \( \boxed{ n = 1 } \) for part (b).