Find the sum of all possible $n$ such that $n$ is a positive integer and there exist $a, b, c$ real numbers such that for every integer $m$, the quantity $\frac{2013m^3 + am^2 + bm + c}{n}$ is an integer.
1. **Initial Condition Analysis:**
   Given the expression \(\frac{2013m^3 + am^2 + bm + c}{n}\) must be an integer for every integer \(m\), we start by considering \(m = 0\). This gives:
   \[
   \frac{c}{n} \text{ must be an integer, implying } n \text{ divides } c.
   \]

2. **Polynomial Modulo \(n\):**
   For any integer \(m\), the expression \(2013m^3 + am^2 + bm + c\) must be divisible by \(n\). This implies:
   \[
   2013m^3 + am^2 + bm + c \equiv 0 \pmod{n}.
   \]

3. **Considering \(m \neq 0\):**
   For \(m \neq 0\), we need:
   \[
   2013m^3 + am^2 + bm \equiv 0 \pmod{n}.
   \]
   This can be rewritten as:
   \[
   m(2013m^2 + am + b) \equiv 0 \pmod{n}.
   \]
   Since this must hold for all \(m\), it implies:
   \[
   2013m^2 + am + b \equiv 0 \pmod{n}.
   \]

4. **Substitution and Simplification:**
   Replace \(m\) by \(m + k\) and subtract the original equation:
   \[
   2013(m+k)^2 + a(m+k) + b \equiv 0 \pmod{n},
   \]
   \[
   2013m^2 + 2 \cdot 2013mk + 2013k^2 + am + ak + b \equiv 0 \pmod{n}.
   \]
   Subtracting \(2013m^2 + am + b \equiv 0 \pmod{n}\) from this, we get:
   \[
   2 \cdot 2013mk + 2013k^2 + ak \equiv 0 \pmod{n}.
   \]
   Since this must hold for all \(m\) and \(k\), we can set \(k = 1\):
   \[
   2 \cdot 2013m + 2013 + a \equiv 0 \pmod{n}.
   \]
   This implies:
   \[
   2 \cdot 2013m + 2013 + a \equiv 0 \pmod{n}.
   \]
   For this to hold for all \(m\), the coefficients of \(m\) must be zero modulo \(n\):
   \[
   2 \cdot 2013 \equiv 0 \pmod{n} \quad \text{and} \quad 2013 + a \equiv 0 \pmod{n}.
   \]

5. **Divisibility Conditions:**
   From \(2 \cdot 2013 \equiv 0 \pmod{n}\), we get:
   \[
   n \text{ must divide } 2 \cdot 2013 = 4026.
   \]
   The prime factorization of 4026 is:
   \[
   4026 = 2 \times 3 \times 11 \times 61.
   \]
   Therefore, the possible values of \(n\) are the divisors of 4026.

6. **Sum of Divisors:**
   The sum of the divisors of 4026 can be calculated using the formula for the sum of divisors:
   \[
   \sigma(4026) = (1 + 2)(1 + 3)(1 + 11)(1 + 61).
   \]
   Calculating each term:
   \[
   (1 + 2) = 3, \quad (1 + 3) = 4, \quad (1 + 11) = 12, \quad (1 + 61) = 62.
   \]
   Therefore:
   \[
   \sigma(4026) = 3 \times 4 \times 12 \times 62 = 2976.
   \]

Conclusion:
\[
\boxed{2976}
\]