A positive interger number $k$ is called “$t-m$”-property if forall positive interger number $a$, there exists a positive integer number $n$ such that
${{1}^{k}}+{{2}^{k}}+{{3}^{k}}+...+{{n}^{k}} \equiv a (\bmod m).$
a) Find all positive integer numbers $k$ which has $t-20$-property.
b) Find smallest positive integer number $k$ which has $t-{{20}^{15}}$-property.

Let's break down the solution into detailed steps for both parts (a) and (b).

### Part (a): Finding all positive integer numbers \( k \) which have the \( t-20 \)-property

1. **Definition and Key Lemma**:
   - A positive integer \( k \) has the \( t-20 \)-property if for all positive integers \( a \), there exists a positive integer \( n \) such that:
     \[
     1^k + 2^k + 3^k + \cdots + n^k \equiv a \pmod{20}
     \]
   - Key Lemma: \( 1^k + 2^k + \cdots + (p-1)^k \equiv 0 \pmod{p} \) if \( k \not\equiv 0 \pmod{p-1} \).

2. **Application of the Key Lemma**:
   - For \( p = 5 \), the lemma implies that \( 1^k + 2^k + \cdots + 4^k \equiv 0 \pmod{5} \) if \( k \not\equiv 0 \pmod{4} \).
   - For \( p = 4 \), the lemma implies that \( 1^k + 2^k + 3^k \equiv 0 \pmod{4} \) if \( k \not\equiv 0 \pmod{3} \).

3. **Conclusion for \( t-20 \)-property**:
   - Since \( 20 = 4 \times 5 \), \( k \) must satisfy both \( k \equiv 0 \pmod{4} \) and \( k \equiv 0 \pmod{3} \).
   - Therefore, \( k \) must be a multiple of the least common multiple of 4 and 3, which is 12.
   - Hence, all positive integers \( k \) that have the \( t-20 \)-property are multiples of 12.

### Part (b): Finding the smallest positive integer number \( k \) which has the \( t-20^{15} \)-property

1. **Claim**:
   - The smallest positive integer \( k \) which has the \( t-20^{15} \)-property is \( k = 4 \).

2. **Proof for \( k \leq 3 \) fails**:
   - Using the Key Lemma, for \( p = 5 \), \( 1^k + 2^k + \cdots + 4^k \equiv 0 \pmod{5} \) if \( k \not\equiv 0 \pmod{4} \).
   - For \( k \leq 3 \), \( k \not\equiv 0 \pmod{4} \), so \( \sum_{i=1}^n i^k \) does not form a complete residue set modulo 5.
   - Therefore, \( k \leq 3 \) fails to have the \( t-20^{15} \)-property.

3. **Proof for \( k = 4 \) works**:
   - We need to show that \( \sum_{i=1}^n i^4 \) forms a complete residue set modulo \( 20^l \) for all \( l \).
   - Check that \( \sum_{i=1}^n i^4 = \frac{6n^5 + 15n^4 + 10n^2 - n}{30} \) forms a complete residue set modulo 20.
   - Use induction to show that if it works for \( l \), it works for \( l+1 \).

4. **Induction Step**:
   - Assume \( X \equiv \frac{6n^5 + 15n^4 + 10n^2 - n}{30} \pmod{20^l} \) with \( 0 \leq X < 20^l \).
   - Find \( m_i \) such that:
     \[
     X + i \cdot 20^l \equiv \frac{6m_i^5 + 15m_i^4 + 10m_i^2 - m_i}{30} \pmod{20^{l+1}}
     \]
   - Simplify using \( f(n) = 6n^5 + 15n^4 + 10n^2 - n \):
     \[
     f(n + 60^l \cdot t) \equiv f(n) + (30n^4 + 60n^3 + 20n - 1) \cdot 60^l \cdot t \pmod{30 \cdot 20^{l+1}}
     \]
   - Find \( t \) such that:
     \[
     (30n^4 + 60n^3 + 20n - 1) \cdot 60^l \cdot t \equiv i \cdot 30 \cdot 20^l \pmod{30 \cdot 20^{l+1}}
     \]
   - Reduce to:
     \[
     (30n^4 + 60n^3 + 20n - 1) \cdot 3^{l-1} \cdot t \equiv 10i \pmod{200}
     \]
   - Take \( t = 10t' \), then:
     \[
     (30n^4 + 60n^3 + 20n - 1) \cdot 3^{l-1} \cdot t' \equiv i \pmod{20}
     \]
   - Since \( (30n^4 + 60n^3 + 20n - 1, 20) = 1 \) and \( (3^{l-1}, 20) = 1 \), for each \( i \), we can find a suitable \( t' \).

5. **Conclusion**:
   - The induction is complete, and \( k = 4 \) works for the \( t-20^{15} \)-property.

The final answer is \( \boxed{ k = 4 } \).