What is the smallest positive integer $n$ such that $2013^n$ ends in $001$ (i.e. the rightmost three digits of $2013^n$ are $001$?
1. We need to find the smallest positive integer \( n \) such that \( 2013^n \) ends in \( 001 \). This can be written as:
   \[
   2013^n \equiv 1 \pmod{1000}
   \]

2. Using the Chinese Remainder Theorem, we can break this down into two congruences:
   \[
   2013^n \equiv 1 \pmod{8} \quad \text{and} \quad 2013^n \equiv 1 \pmod{125}
   \]

3. First, consider the congruence modulo 8:
   \[
   2013 \equiv 5 \pmod{8}
   \]
   Thus, we need:
   \[
   5^n \equiv 1 \pmod{8}
   \]
   The order of 5 modulo 8 is 2, because:
   \[
   5^2 = 25 \equiv 1 \pmod{8}
   \]
   Therefore, \( n \) must be even.

4. Next, consider the congruence modulo 125:
   \[
   2013 \equiv 13 \pmod{125}
   \]
   Thus, we need:
   \[
   13^n \equiv 1 \pmod{125}
   \]
   The order of 13 modulo 125 can be found using Euler's theorem, which states that \( a^{\phi(n)} \equiv 1 \pmod{n} \) for \( \gcd(a, n) = 1 \). Here, \( \phi(125) = 100 \), so:
   \[
   13^{100} \equiv 1 \pmod{125}
   \]
   Therefore, \( n \) must be a divisor of 100.

5. Since \( n \) must be even and a divisor of 100, the possible values for \( n \) are 2, 4, 10, 20, 50, and 100. We need to check these values to find the smallest \( n \) such that:
   \[
   13^n \equiv 1 \pmod{125}
   \]

6. Checking \( n = 2 \):
   \[
   13^2 = 169 \equiv 44 \pmod{125}
   \]
   This does not work.

7. Checking \( n = 4 \):
   \[
   13^4 = (13^2)^2 = 44^2 = 1936 \equiv 61 \pmod{125}
   \]
   This does not work.

8. Checking \( n = 10 \):
   \[
   13^{10} = (13^5)^2 \equiv 18^2 = 324 \equiv 74 \pmod{125}
   \]
   This does not work.

9. Checking \( n = 20 \):
   \[
   13^{20} = (13^{10})^2 \equiv 74^2 = 5476 \equiv 51 \pmod{125}
   \]
   This does not work.

10. Checking \( n = 50 \):
    \[
    13^{50} = (13^{25})^2 \equiv 18^2 = 324 \equiv 74 \pmod{125}
    \]
    This does not work.

11. Checking \( n = 100 \):
    \[
    13^{100} \equiv 1 \pmod{125}
    \]
    This works.

Therefore, the smallest positive integer \( n \) such that \( 2013^n \) ends in \( 001 \) is \( \boxed{100} \).