Let $f$ be a non-constant polynomial such that \[ f(x-1) + f(x) + f(x+1) = \frac {f(x)^2}{2013x} \] for all nonzero real numbers $x$. Find the sum of all possible values of $f(1)$.

[i]Proposed by Ahaan S. Rungta[/i]
1. Let \( f \) be a polynomial of degree \( d \). We start by comparing the degrees of both sides of the given equation:
   \[
   f(x-1) + f(x) + f(x+1) = \frac{f(x)^2}{2013x}
   \]
   The left-hand side (LHS) is a sum of three polynomials each of degree \( d \), so the degree of the LHS is \( d \).

2. The right-hand side (RHS) is a fraction where the numerator is \( f(x)^2 \), which has degree \( 2d \), and the denominator is \( 2013x \), which has degree 1. Therefore, the degree of the RHS is:
   \[
   2d - 1
   \]

3. Equating the degrees of both sides, we get:
   \[
   d = 2d - 1
   \]
   Solving for \( d \):
   \[
   d - 2d = -1 \implies -d = -1 \implies d = 1
   \]
   Thus, \( f \) is a linear polynomial.

4. Let \( f(x) = mx + b \). Substituting \( f(x) = mx + b \) into the original equation, we get:
   \[
   f(x-1) + f(x) + f(x+1) = \frac{(mx + b)^2}{2013x}
   \]
   Calculating \( f(x-1) \), \( f(x) \), and \( f(x+1) \):
   \[
   f(x-1) = m(x-1) + b = mx - m + b
   \]
   \[
   f(x) = mx + b
   \]
   \[
   f(x+1) = m(x+1) + b = mx + m + b
   \]
   Adding these, we get:
   \[
   f(x-1) + f(x) + f(x+1) = (mx - m + b) + (mx + b) + (mx + m + b) = 3mx + 3b
   \]

5. Substituting into the original equation:
   \[
   3mx + 3b = \frac{(mx + b)^2}{2013x}
   \]
   Simplifying the RHS:
   \[
   \frac{(mx + b)^2}{2013x} = \frac{m^2x^2 + 2mbx + b^2}{2013x} = \frac{m^2x^2}{2013x} + \frac{2mbx}{2013x} + \frac{b^2}{2013x} = \frac{m^2x}{2013} + \frac{2mb}{2013} + \frac{b^2}{2013x}
   \]

6. For the degrees to match, the term \(\frac{b^2}{2013x}\) must be zero, implying \( b = 0 \). Thus, \( f(x) = mx \).

7. Substituting \( f(x) = mx \) into the equation:
   \[
   3mx = \frac{m^2x^2}{2013x} = \frac{m^2x}{2013}
   \]
   Simplifying:
   \[
   3m = \frac{m^2}{2013}
   \]
   Solving for \( m \):
   \[
   3m \cdot 2013 = m^2 \implies 3 \cdot 2013 = m \implies m = 6039
   \]

8. Therefore, \( f(x) = 6039x \) and \( f(1) = 6039 \).

The final answer is \( \boxed{6039} \)