For a positive integer number $r$, the sequence $a_{1},a_{2},...$ defined by $a_{1}=1$ and $a_{n+1}=\frac{na_{n}+2(n+1)^{2r}}{n+2}\forall n\geq 1$. Prove that each $a_{n}$ is positive integer number, and find $n's$ for which $a_{n}$ is even.
To prove that each \( a_n \) is a positive integer and to find the values of \( n \) for which \( a_n \) is even, we will proceed as follows:

1. **Define the sequence and initial conditions:**
   \[
   a_1 = 1
   \]
   \[
   a_{n+1} = \frac{n a_n + 2(n+1)^{2r}}{n+2} \quad \forall n \geq 1
   \]

2. **Inductive Hypothesis:**
   Assume \( a_n \) is a positive integer for some \( n \geq 1 \). We need to show that \( a_{n+1} \) is also a positive integer.

3. **Inductive Step:**
   Consider the expression for \( a_{n+1} \):
   \[
   a_{n+1} = \frac{n a_n + 2(n+1)^{2r}}{n+2}
   \]
   We need to show that the numerator \( n a_n + 2(n+1)^{2r} \) is divisible by \( n+2 \).

4. **Divisibility Check:**
   Let \( S_n = \sum_{k=1}^n k^{2r+1} \). We claim that:
   \[
   a_n = \frac{2}{n(n+1)} S_n
   \]
   We will prove this by induction.

   - **Base Case:**
     For \( n = 1 \):
     \[
     a_1 = 1 = \frac{2}{1 \cdot 2} \cdot 1^{2r+1} = \frac{2}{2} = 1
     \]
     which is true.

   - **Inductive Step:**
     Assume \( a_n = \frac{2}{n(n+1)} S_n \) holds for some \( n \geq 1 \). We need to show it holds for \( n+1 \):
     \[
     a_{n+1} = \frac{n a_n + 2(n+1)^{2r}}{n+2}
     \]
     Substitute \( a_n = \frac{2}{n(n+1)} S_n \):
     \[
     a_{n+1} = \frac{n \left( \frac{2}{n(n+1)} S_n \right) + 2(n+1)^{2r}}{n+2}
     \]
     Simplify the expression:
     \[
     a_{n+1} = \frac{2 S_n + 2 n (n+1)^{2r}}{n(n+1)(n+2)}
     \]
     \[
     a_{n+1} = \frac{2 (S_n + (n+1)^{2r})}{(n+1)(n+2)}
     \]
     Notice that \( S_{n+1} = S_n + (n+1)^{2r+1} \):
     \[
     a_{n+1} = \frac{2 S_{n+1}}{(n+1)(n+2)}
     \]
     This completes the inductive step, proving that \( a_n = \frac{2}{n(n+1)} S_n \) for all \( n \geq 1 \).

5. **Positivity:**
   Since \( S_n \) is a sum of positive terms, \( a_n \) is positive for all \( n \).

6. **Evenness:**
   To determine when \( a_n \) is even, consider the parity of \( S_n \):
   - For \( n \equiv 0, 3 \pmod{4} \), \( S_n \) is even, making \( a_n \) even.
   - For \( n \equiv 1, 2 \pmod{4} \), \( S_n \) is odd, making \( a_n \) odd.

   Therefore, \( a_n \) is even if and only if \( n \equiv 0 \) or \( 3 \pmod{4} \).

The final answer is \( \boxed{ n \equiv 0 } \) or \( 3 \pmod{4} \).