For all positive integer $n$, we consider the number $$a_n =4^{6^n}+1943$$ Prove that $a_n$ is dividible by $2013$ for all $n\ge 1$, and find all values of $n$ for which $a_n - 207$ is the cube of a positive integer.
1. **Prove that \( a_n \) is divisible by 2013 for all \( n \geq 1 \):**

   We start by noting that \( 2013 = 3 \times 11 \times 61 \). We need to show that \( 4^{6^n} + 1943 \equiv 0 \pmod{2013} \).

   - **Modulo 3:**
     \[
     4 \equiv 1 \pmod{3} \implies 4^{6^n} \equiv 1^{6^n} \equiv 1 \pmod{3}
     \]
     Therefore,
     \[
     4^{6^n} + 1943 \equiv 1 + 1943 \equiv 1944 \equiv 0 \pmod{3}
     \]

   - **Modulo 11:**
     \[
     4^5 \equiv 1 \pmod{11} \implies 4^{6^n} \equiv 4^{6^n \mod 5} \pmod{11}
     \]
     Since \( 6^n \mod 5 \) cycles through 1, 2, 4, 3, 1, ..., we have:
     \[
     4^{6^n} \equiv 4^1 \equiv 4 \pmod{11}
     \]
     Therefore,
     \[
     4^{6^n} + 1943 \equiv 4 + 1943 \equiv 4 + 7 \equiv 11 \equiv 0 \pmod{11}
     \]

   - **Modulo 61:**
     \[
     4^{60} \equiv 1 \pmod{61} \implies 4^{6^n} \equiv 4^{6^n \mod 60} \pmod{61}
     \]
     Since \( 6^n \mod 60 \) cycles through 6, 36, 16, 36, ..., we have:
     \[
     4^{6^n} \equiv 4^6 \equiv 4096 \equiv 70 \pmod{61}
     \]
     Therefore,
     \[
     4^{6^n} + 1943 \equiv 70 + 1943 \equiv 2013 \equiv 0 \pmod{61}
     \]

   Since \( 4^{6^n} + 1943 \equiv 0 \pmod{3} \), \( 4^{6^n} + 1943 \equiv 0 \pmod{11} \), and \( 4^{6^n} + 1943 \equiv 0 \pmod{61} \), by the Chinese Remainder Theorem, we have:
   \[
   4^{6^n} + 1943 \equiv 0 \pmod{2013}
   \]

2. **Find all values of \( n \) for which \( a_n - 207 \) is the cube of a positive integer:**

   We need to check when \( a_n - 207 \) is a perfect cube. We have:
   \[
   a_n - 207 = 4^{6^n} + 1943 - 207 = 4^{6^n} + 1736
   \]

   - For \( n = 1 \):
     \[
     a_1 - 207 = 4^{6^1} + 1736 = 4^6 + 1736 = 4096 + 1736 = 5832
     \]
     Since \( 5832 = 18^3 \), \( a_1 - 207 \) is a perfect cube.

   - For \( n \geq 2 \):
     \[
     4^{6^n} \equiv 1 \pmod{9} \implies a_n - 207 \equiv 1 + 1736 \equiv 1737 \equiv 0 \pmod{9}
     \]
     However,
     \[
     4^{6^n} \equiv 1 \pmod{27} \implies a_n - 207 \equiv 1 + 1736 \equiv 1737 \equiv 9 \pmod{27}
     \]
     Therefore, \( v_3(a_n - 207) = 2 \), which means \( a_n - 207 \) is not a cube for \( n \geq 2 \).

   Hence, the only value of \( n \) for which \( a_n - 207 \) is a perfect cube is \( n = 1 \).

The final answer is \( \boxed{ n = 1 } \).