Find the smallest positive integer $n$ that satisfies the following:

We can color each positive integer with one of $n$ colors such that the equation $w + 6x = 2y + 3z$ has no solutions in positive integers with all of $w, x, y$ and $z$ having the same color. (Note that $w, x, y$ and $z$ need not be distinct.)
1. **Prove that \( n > 3 \):**
   - Assume for the sake of contradiction that there are at most 3 colors, say \( c_1 \), \( c_2 \), and \( c_3 \).
   - Let 1 have color \( c_1 \) without loss of generality.
   - By considering the tuple \((1,1,2,1)\), the color of 2 must be different from \( c_1 \), so let it be \( c_2 \).
   - By considering the tuple \((3,2,3,3)\), the color of 3 cannot be \( c_2 \), and by \((3,1,3,1)\), the color of 3 cannot be \( c_1 \). Therefore, the color of 3 must be \( c_3 \).
   - By considering the tuple \((6,2,6,2)\), the color of 6 cannot be \( c_2 \), and by \((3,3,6,3)\), the color of 6 cannot be \( c_3 \). Therefore, the color of 6 must be \( c_1 \).
   - By considering the tuple \((9,6,9,9)\), the color of 9 cannot be \( c_1 \), and by \((9,3,9,3)\), the color of 9 cannot be \( c_3 \). Therefore, the color of 9 must be \( c_2 \).
   - By considering the tuple \((6,4,6,6)\), the color of 4 cannot be \( c_1 \), and by \((2,2,4,2)\), the color of 4 cannot be \( c_2 \). Therefore, the color of 4 must be \( c_3 \).
   - By considering the tuple \((6,6,12,6)\), the color of 12 cannot be \( c_1 \), and by \((12,4,12,4)\), the color of 12 cannot be \( c_3 \). Therefore, the color of 12 must be \( c_2 \).
   - However, we have reached a contradiction, as \((12,2,9,2)\) is colored with only \( c_2 \). Therefore, there must be more than 3 colors.

2. **Prove that \( n = 4 \) works:**
   - Define the following four sets:
     \[
     \begin{align*}
     c_1 &= \{3^{2a}(3b+1) \mid a, b \ge 0\} \\
     c_2 &= \{3^{2a}(3b+2) \mid a, b \ge 0\} \\
     c_3 &= \{3^{2a+1}(3b+1) \mid a, b \ge 0\} \\
     c_4 &= \{3^{2a+1}(3b+2) \mid a, b \ge 0\}
     \end{align*}
     \]
   - It is obvious that each positive integer appears in exactly one of these sets because if we divide out all powers of 3 in a number, then we will get a number that is either 1 or 2 modulo 3.
   - We assert that no quadruple of positive integers \((w, x, y, z)\) satisfying \( w + 6x = 2y + 3z \) consists of four members from the same set.
   - Assume for the sake of contradiction that \((a, b, c, d)\) are positive integers satisfying \( a + 6b = 2c + 3d \) with \( a, b, c, d \) from the same set.
   - If \((a, b, c, d)\) are from the same set, then \(\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}, \frac{d}{3}\right)\) are also from the same set. Hence, we may assume without loss of generality that at least one of \((a, b, c, d)\) is not divisible by 3, or else we can perform this operation to reduce it.
   - Let \( v_3(t) \) denote the largest integer such that \( 3^{v_3(t)} \mid t \). Since at least one of these is not divisible by 3, \( v_3(a), v_3(b), v_3(c), v_3(d) \) must be even.
   - Notice that \( w + 6x \equiv w \equiv 2y + 3z \equiv -y \pmod{3} \), so \( w + y \equiv 0 \pmod{3} \). Therefore, \( w \equiv y \equiv 0 \pmod{3} \), or else they would have different residues modulo 3. Also, they must both be divisible by 9 because \( v_3(w) \) and \( v_3(y) \) are even.
   - Since \( 2y - w = 3(2x - z) \), \( 3 \mid 2x - z \implies x + z \equiv 0 \pmod{3} \). However, this implies that they are both divisible by 3, or else they would have different residues modulo 3. This is a contradiction, as we have assumed that at least one of \( w, x, y, z \) is not divisible by 3. Therefore, this coloring works.

Thus, \( n = \boxed{4} \) is our answer.