Source: 2017 Canadian Open Math Challenge, Problem B4
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Numbers $a$, $b$ and $c$ form an arithmetic sequence if $b - a = c - b$. Let $a$, $b$, $c$ be positive integers forming an arithmetic sequence with $a < b < c$. Let $f(x) = ax2 + bx + c$. Two distinct real numbers $r$ and $s$ satisfy $f(r) = s$ and $f(s) = r$. If $rs = 2017$, determine the smallest possible value of $a$.
1. Given that \(a\), \(b\), and \(c\) form an arithmetic sequence, we have:
   \[
   b - a = c - b \implies 2b = a + c \implies c = 2b - a
   \]
   Since \(a < b < c\), we know \(a\), \(b\), and \(c\) are positive integers.

2. The function \(f(x) = ax^2 + bx + c\) is given, and we know that \(f(r) = s\) and \(f(s) = r\) for distinct real numbers \(r\) and \(s\). Additionally, \(rs = 2017\).

3. Since \(f(r) = s\) and \(f(s) = r\), we can write:
   \[
   f(x) = k(x - r)(x - s) - x + (r + s)
   \]
   for some constant \(k\). Expanding this, we get:
   \[
   f(x) = k(x^2 - (r + s)x + rs) - x + (r + s)
   \]
   Given \(rs = 2017\), this simplifies to:
   \[
   f(x) = kx^2 - k(r + s)x + 2017k - x + (r + s)
   \]

4. Comparing this with the standard form \(f(x) = ax^2 + bx + c\), we identify:
   \[
   a = k, \quad b = -k(r + s) - 1, \quad c = 2017k + (r + s)
   \]

5. To find \(k\), we use the fact that \(f(x)\) must be a quadratic polynomial. We equate the coefficients:
   \[
   2018k + (r + s) = 2k(r + s)
   \]
   Solving for \(k\), we get:
   \[
   2018k + (r + s) = 2k(r + s) \implies 2018k = k(r + s) - (r + s) \implies k = \frac{r + s}{2(r + s - 2018)}
   \]

6. For \(k\) to be an integer, \(2(r + s - 2018)\) must divide \(r + s\). This implies:
   \[
   r + s - 2018 \mid 2018
   \]

7. To minimize \(a = k\), we need to maximize \(r + s\). The divisors of 2018 are \(1, 2, 1009, 2018\). The maximum value of \(r + s\) occurs when:
   \[
   r + s - 2018 = 1 \implies r + s = 2019
   \]

8. Substituting \(r + s = 4036\) into the expression for \(k\), we get:
   \[
   k = \frac{4036}{2(4036 - 2018)} = \frac{4036}{2 \times 2018} = 1
   \]

9. Therefore, the smallest possible value of \(a\) is:
   \[
   a = k = 1
   \]

The final answer is \(\boxed{1}\).