For any positive integers $a,b,c,n$, we define 
$$D_n(a,b,c)=\mathrm{gcd}(a+b+c,a^2+b^2+c^2,a^n+b^n+c^n).$$

1. Prove that if $n$ is a positive integer not divisible by $3$, then for any positive integer $k$, there exist three integers $a,b,c$ such that $\mathrm{gcd}(a,b,c)=1$, and $D_n(a,b,c)>k$. 

2. For any positive integer $n$ divisible by $3$, find all values of $D_n(a,b,c)$, where $a,b,c$ are three positive integers such that $\mathrm{gcd}(a,b,c)=1$. 
### Part 1:
1. Let \( p \equiv 1 \pmod{6} \) be a prime. We claim:
   - (a) There exist \( a, b, c \) such that \( \gcd(a, b, c) = 1 \), \( a^3 \equiv b^3 \equiv c^3 \pmod{p} \), and \( a, b, c \) are distinct modulo \( p \).
   - (b) \( p \mid D_n(a, b, c) \) for any such triple.

2. To show (a), let \( k \) be a cubic root modulo \( p \), and let \( g \) be a primitive root such that \( g^{3i} \equiv k \pmod{p} \) for some \( i \). Define:
   \[
   a \equiv g^i \pmod{p}, \quad b \equiv g^{i + \frac{p-1}{3}} \pmod{p}, \quad c \equiv g^{i + 2\frac{p-1}{3}} \pmod{p}.
   \]
   Clearly, \( a^3 \equiv b^3 \equiv c^3 \pmod{p} \) and they are distinct since \( g \) is a primitive root. To ensure \( \gcd(a, b, c) = 1 \), select \( c \equiv 1 \pmod{a} \). This ensures \( \gcd(a, b, c) \mid \gcd(a, c) = 1 \).

3. To show (b), note that if \( a^3 \equiv b^3 \equiv c^3 \pmod{p} \), then \( p \mid (a-b)(a^2+ab+b^2) \). Since \( p \nmid a-b \), we get \( p \mid a^2+ab+b^2 \). Similarly, \( p \mid b^2+bc+c^2 \) and \( p \mid c^2+ca+a^2 \). Subtracting the second from the first, we get \( p \mid (a-c)(a+b+c) \), yielding \( p \mid a+b+c \).

4. Summing the equations, we get \( p \mid 3(a^2+b^2+c^2) + (ab+bc+ca) \). Together with \( p \mid 3(a+b+c)^2 = 3(a^2+b^2+c^2) + 6(ab+bc+ca) \), we get \( p \mid 5(ab+bc+ca) \), hence \( p \mid ab+bc+ca \). Therefore, \( p \mid a^2+b^2+c^2 \).

5. Finally, if \( n = 3\ell + 1 \) and \( a^3 \equiv b^3 \equiv c^3 \equiv k \pmod{p} \), then \( a^n + b^n + c^n \equiv k^\ell(a+b+c) \equiv 0 \pmod{p} \). Similarly for \( n = 3\ell + 2 \). This establishes Claim (b), thus part 1.

### Part 2:
1. Note that \( D_n \mid (a+b+c)^2 - (a^2+b^2+c^2) \), yielding \( D_n \mid 2(ab+bc+ca) \). If \( 2 \mid D_n \), then two of \( a, b, c \) are odd and one is even, thus \( a^2 + b^2 + c^2 \equiv 2 \pmod{4} \). Consequently, \( v_2(D_n) \le 1 \).

2. Take any odd divisor \( d \) of \( D_n \). Then \( d \mid ab+bc+ca \). Using \( c \equiv -(a+b) \pmod{d} \), we get \( d \mid a^2 + ab + b^2 \). This yields \( a^3 \equiv b^3 \equiv c^3 \equiv k \pmod{d} \). Note also that \( \gcd(d, a) = \gcd(d, b) = \gcd(d, c) = 1 \) as otherwise if \( q \mid (d, a) \) a prime, then \( q \mid b, c \), contradiction.

3. Finally, as \( n = 3\ell \) for some \( \ell \), we get \( a^n + b^n + c^n \equiv 3k^\ell \pmod{d} \), thus \( d \mid 3 \). From here, we get \( D_n \mid 6 \). It is easy to see \( D_n(3, 1, 1) = 1 \), \( D_n(1, 1, 1) = 3 \), \( D_n(2, 1, 1) = 2 \), and \( D_n(4, 1, 1) = 6 \).

The final answer is \( \boxed{ D_n \mid 6 } \).