Find all positive integers $k$ for which there exists a nonlinear function $f:\mathbb{Z} \rightarrow\mathbb{Z}$ such that the equation $$f(a)+f(b)+f(c)=\frac{f(a-b)+f(b-c)+f(c-a)}{k}$$ holds for any integers $a,b,c$ satisfying $a+b+c=0$ (not necessarily distinct).

[i]Evan Chen[/i]
We need to find all positive integers \( k \) for which there exists a nonlinear function \( f: \mathbb{Z} \rightarrow \mathbb{Z} \) such that the equation
\[ f(a) + f(b) + f(c) = \frac{f(a-b) + f(b-c) + f(c-a)}{k} \]
holds for any integers \( a, b, c \) satisfying \( a + b + c = 0 \).

1. **Verification for \( k = 0 \):**
   Suppose \( k = 0 \). We claim the function
   \[
   f(x) = \begin{cases} 
   0 & \text{if } x \equiv 0 \pmod{3}, \\
   |x| & \text{if } |x| \equiv 1 \pmod{3}, \\
   -|x| & \text{if } |x| \equiv 2 \pmod{3}
   \end{cases}
   \]
   works. Clearly, \( f \) is nonlinear. Note that \( a - b, b - c, c - a \) are congruent modulo 3. Let \( x \) and \( y \) be two of \( a - b, b - c, c - a \) with the same sign. Then the third equals \( -(x + y) \). We need to show
   \[
   f(|x|) + f(|y|) + f(|x| + |y|) = 0.
   \]
   We consider three cases:
   - \( |x| \equiv |y| \equiv 0 \pmod{3} \): \( f(|x|) = f(|y|) = f(|x| + |y|) = 0 \).
   - \( |x| \equiv |y| \equiv 1 \pmod{3} \): \( |x| + |y| \equiv 2 \pmod{3} \). Then \( f(|x|) = |x| \), \( f(|y|) = |y| \), and \( f(|x| + |y|) = -(|x| + |y|) \).
   - \( |x| \equiv |y| \equiv 2 \pmod{3} \): \( |x| + |y| \equiv 1 \pmod{3} \). Then \( f(|x|) = -|x| \), \( f(|y|) = -|y| \), and \( f(|x| + |y|) = |x| + |y| \).

   In all three cases, \( f(|x|) + f(|y|) + f(|x| + |y|) = 0 \).

2. **Verification for \( k = 1 \):**
   Suppose \( k = 1 \). We claim the function
   \[
   f(x) = \begin{cases} 
   0 & \text{if } x \text{ is even}, \\
   1 & \text{if } x \text{ is odd}
   \end{cases}
   \]
   works. Clearly, \( f \) is nonlinear. We consider two cases:
   - All \( a, b, c \) are even: Both sides of the equation equal 0.
   - Two of \( a, b, c \) are odd and the third is even: The left-hand side equals 2. Among \( a - b, b - c, c - a \), two numbers are odd and one is even, so the right-hand side also equals 2.

3. **Verification for \( k = 3 \):**
   Suppose \( k = 3 \). We claim the function \( f(x) = x^2 \) works. Clearly, \( f \) is nonlinear. Verification:
   \[
   f(a-b) + f(b-c) + f(c-a) = (a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 + b^2 + c^2) - (2ab + 2bc + 2ca) = 3(a^2 + b^2 + c^2) = 3(f(a) + f(b) + f(c)).
   \]

4. **Verification for \( k = 9 \):**
   Suppose \( k = 9 \). We claim the function \( f(x) = x^4 \) works. Clearly, \( f \) is nonlinear. Verification:
   \[
   f(a-b) + f(b-c) + f(c-a) = (a-b)^4 + (b-c)^4 + (c-a)^4 = 9(a^4 + b^4 + c^4) = 9(f(a) + f(b) + f(c)).
   \]

5. **Proof that these are the only values:**
   Suppose such a function \( f \) exists for some \( k \ne 0, 1, 3, 9 \). We will show a contradiction.

   - **Claim 1:** \( f(a-b) + f(b-c) + f(c-a) = f(b-a) + f(c-b) + f(a-c) \) for all \( a + b + c = 0 \).
     *Proof:* Compare \( P(a, b, c) \) and \( P(a, c, b) \).

   - **Claim 2:** \( f(a) + f(b) + f(c) = f(-a) + f(-b) + f(-c) \) for all \( a + b + c = 0 \).
     *Proof:* Compare \( P(a, b, c) \) and \( P(-a, -b, -c) \), and use Claim 1.

   - **Claim 3:** Define \( g(x) = f(x) - f(-x) \). Then \( g(x) \equiv mx \).
     *Proof:* From Claim 2, we have \( g(a) + g(b) + g(c) = 0 \) for all \( a + b + c = 0 \). Fix \( a \) and vary \( b, c \). We get \( g(b_1) + g(c_1) = g(b_2) + g(c_2) \) for all \( b_1 + c_1 = b_2 + c_2 \). Setting \( c_1 = 0 \), we get \( g(x + y) = g(x) + g(y) \) for all \( x, y \). This is the Cauchy equation, and since \( g: \mathbb{Z} \to \mathbb{Z} \), \( g(x) \equiv mx \).

   - **Claim 4:** There exists an even function \( h \) satisfying the required conditions.
     *Proof:* Define \( h(x) = 2f(x) - mx \). Clearly, \( h(x) \in \mathbb{Z} \), \( h \) is even, and \( h \) satisfies \( P(a, b, c) \). Further, \( h \) is nonlinear as \( f \) is nonlinear.

   - **Corollary 5:** We may assume \( f \) is even and then prove a contradiction.
     *Proof:* We can work with \( h \) instead of \( f \).

   Now, assume \( f \) is even. Since \( k \ne 1 \), \( P(0, 0, 0) \) gives \( f(0) = 0 \). Look at \( P(x, -x, 0) \):
   \[
   f(x) + f(x) + f(0) = \frac{f(2x) + f(x) + f(x)}{k} \implies f(2x) = (2k - 2)f(x).
   \]
   Look at \( P(x, x, -2x) \):
   \[
   f(x) + f(x) + f(2x) = \frac{f(0) + f(3x) + f(3x)}{k} \implies f(3x) = k^2 f(x).
   \]

   - **Claim 6:** There exist polynomials \( \{p_n(k)\}_{n \in \mathbb{Z}} \) such that \( f(nx) = p_n(k) \cdot f(x) \) for all \( n, k \in \mathbb{Z} \).
     *Proof:* Fix \( x \). As \( f \) is even, it suffices to prove the claim for \( n \ge 0 \). Use strong induction on \( m \ge 1 \) to show the assertion is true for all \( n = 0, 1, 2, \ldots, 2m-1 \). Base cases \( m = 1, 2 \) follow from the previous equations. For \( n = 2m \), \( f(2m) = (2k - 2)f(m) \). For \( n = 2m + 1 \), use \( P((m+1)x, (-m)x, x) \).

   - **Claim 7:** We may assume \( f(1) = 1 \).
     *Proof:* By Claim 6, \( f(1) \) divides each \( f(n) \). Change \( f \to \frac{f}{f(1)} \).

   Construct polynomials \( \{p_n(k)\}_{n \in \mathbb{Z}} \) in \( k \) such that \( f(nx) = p_n(k) \cdot f(x) \). Denote by \( L(p) \) the leading coefficient of \( p \). From previous equations, we can take \( p_2(k) = 2k - 2 \), \( p_3(k) = k^2 \), \( p_4(k) = (2k - 2)^2 \), \( p_8(k) = (2k - 2)^3 \). Use \( P(3x, -2x, -x) \) to get \( p_5(k) \) and \( P(4x, -3x, -x) \) to get \( p_7(k) \).

   - **Claim 8:** \( f \) is completely multiplicative, i.e., \( f(mn) = f(m)f(n) \).
     *Proof:* This follows from Claim 6 and \( f(1) = 1 \).

   Finally, look at \( P(5, -3, -2) \):
   \[
   f(5) + f(3) + f(2) = \frac{f(8) + f(1) + f(7)}{k}.
   \]
   This gives a degree four equation in \( k \). Since the equation must be satisfied for \( k = 0, 1, 3, 9 \), and it cannot have any fifth root, we conclude \( k \in \{0, 1, 3, 9\} \).

   \(\blacksquare\)

The final answer is \( \boxed{ k = \{0, 1, 3, 9\} } \).