What is the smallest positive integer $n$ such that $2^n - 1$ is a multiple of $2015$?
1. **Identify the prime factorization of 2015:**
   \[
   2015 = 5 \times 13 \times 31
   \]
   We need to find the smallest positive integer \( n \) such that \( 2^n - 1 \) is divisible by 2015. This means \( 2^n \equiv 1 \pmod{2015} \). By the Chinese Remainder Theorem (CRT), it suffices to find \( n \) such that:
   \[
   2^n \equiv 1 \pmod{5}, \quad 2^n \equiv 1 \pmod{13}, \quad \text{and} \quad 2^n \equiv 1 \pmod{31}
   \]

2. **Apply Fermat's Little Theorem (FLT):**
   Fermat's Little Theorem states that if \( p \) is a prime and \( a \) is an integer not divisible by \( p \), then:
   \[
   a^{p-1} \equiv 1 \pmod{p}
   \]
   For \( p = 5 \):
   \[
   2^4 \equiv 1 \pmod{5}
   \]
   So, the order of 2 modulo 5 divides 4. The possible orders are 1, 2, or 4. Checking these:
   \[
   2^1 \equiv 2 \pmod{5}, \quad 2^2 \equiv 4 \pmod{5}, \quad 2^4 \equiv 1 \pmod{5}
   \]
   Thus, the order of 2 modulo 5 is 4.

   For \( p = 13 \):
   \[
   2^{12} \equiv 1 \pmod{13}
   \]
   So, the order of 2 modulo 13 divides 12. The possible orders are 1, 2, 3, 4, 6, or 12. Checking these:
   \[
   2^1 \equiv 2 \pmod{13}, \quad 2^2 \equiv 4 \pmod{13}, \quad 2^3 \equiv 8 \pmod{13}, \quad 2^4 \equiv 3 \pmod{13}, \quad 2^6 \equiv 12 \equiv -1 \pmod{13}, \quad 2^{12} \equiv 1 \pmod{13}
   \]
   Thus, the order of 2 modulo 13 is 12.

   For \( p = 31 \):
   \[
   2^{30} \equiv 1 \pmod{31}
   \]
   So, the order of 2 modulo 31 divides 30. The possible orders are 1, 2, 3, 5, 6, 10, 15, or 30. Checking these:
   \[
   2^1 \equiv 2 \pmod{31}, \quad 2^2 \equiv 4 \pmod{31}, \quad 2^3 \equiv 8 \pmod{31}, \quad 2^5 \equiv 32 \equiv 1 \pmod{31}, \quad 2^{10} \equiv 1 \pmod{31}, \quad 2^{15} \equiv 1 \pmod{31}, \quad 2^{30} \equiv 1 \pmod{31}
   \]
   Thus, the order of 2 modulo 31 is 30.

3. **Find the least common multiple (LCM):**
   The smallest \( n \) such that \( 2^n \equiv 1 \pmod{5} \), \( 2^n \equiv 1 \pmod{13} \), and \( 2^n \equiv 1 \pmod{31} \) is the least common multiple of the orders found:
   \[
   \text{LCM}(4, 12, 30)
   \]
   Calculate the LCM:
   \[
   \text{LCM}(4, 12) = 12
   \]
   \[
   \text{LCM}(12, 30) = 60
   \]

Conclusion:
\[
\boxed{60}
\]