Find the number of integers $n$ with $n \ge 2$ such that the remainder when $2013$ is divided by $n$ is equal to the remainder when $n$ is divided by $3$.

[i]Proposed by Michael Kural[/i]
We need to find the number of integers \( n \) with \( n \ge 2 \) such that the remainder when \( 2013 \) is divided by \( n \) is equal to the remainder when \( n \) is divided by \( 3 \). This can be expressed as:
\[ 2013 \mod n = n \mod 3. \]

We will consider three cases based on the value of \( n \mod 3 \).

1. **Case 1: \( n \equiv 0 \pmod{3} \)**
   - If \( n \equiv 0 \pmod{3} \), then \( n = 3k \) for some integer \( k \).
   - The condition becomes \( 2013 \mod n = 0 \).
   - This implies \( n \mid 2013 \).
   - The prime factorization of \( 2013 \) is \( 2013 = 3 \times 11 \times 61 \).
   - Therefore, the divisors of \( 2013 \) that are multiples of \( 3 \) are \( 3, 33, 183, \) and \( 2013 \).
   - Thus, there are \( 4 \) solutions in this case.

2. **Case 2: \( n \equiv 1 \pmod{3} \)**
   - If \( n \equiv 1 \pmod{3} \), then \( n = 3k + 1 \) for some integer \( k \).
   - The condition becomes \( 2013 \mod n = 1 \).
   - This implies \( 2012 \equiv 0 \pmod{n} \).
   - Therefore, \( n \mid 2012 \).
   - The prime factorization of \( 2012 \) is \( 2012 = 2^2 \times 503 \).
   - The divisors of \( 2012 \) are \( 1, 2, 4, 503, 1006, \) and \( 2012 \).
   - We need to find the divisors that are \( 1 \pmod{3} \). These are \( 4 \) and \( 1006 \).
   - Thus, there are \( 2 \) solutions in this case.

3. **Case 3: \( n \equiv 2 \pmod{3} \)**
   - If \( n \equiv 2 \pmod{3} \), then \( n = 3k + 2 \) for some integer \( k \).
   - The condition becomes \( 2013 \mod n = 2 \).
   - This implies \( 2011 \equiv 0 \pmod{n} \).
   - Therefore, \( n \mid 2011 \).
   - The prime factorization of \( 2011 \) is \( 2011 \) (since \( 2011 \) is a prime number).
   - The only divisors of \( 2011 \) are \( 1 \) and \( 2011 \).
   - Since \( n \ge 2 \), the only possible value is \( n = 2011 \).
   - Thus, there is \( 1 \) solution in this case.

Adding up all the solutions from the three cases, we get:
\[ 4 + 2 + 1 = 7. \]

The final answer is \(\boxed{7}\).