Correct solutions were submitted by Dan Dima, Al Zimmermann, Jens Voβ, Ahron Teitelman, Juan Carlos Marivela, Lou Cairoli.

The polynomial yields an integer if and only if r is an integer or of the form (9k - 1)/3.

Write r = p/q, with p and q relatively prime.  Substituting, we get the integer 3(p³/q³) + 10(p²/q²) - 3(p/q).  Putting everything over a common denominator, we conclude that q³ divides N = 3p³ + 10p²q - 3pq² and then q divides N, too.  Since q divides the last two summands of N it must divide the first, too.  As p and q are relatively prime, q must divide 3; it follows that either q = 1, and r is itself an integer; or q = 3.  In the latter case 

27 divides  3p³ + 30p² - 27p if and only if 
9 divides  p³ + 10p² - 9p if and only if
9 divides  p³ + 10p²  if and only if
9 divides  p²(p + 10).

Since p is not divisible by 3, 9 can't divide p², and so must divide  p + 10, and  p + 1.  Write p = 9k - 1, for some integer k, then r = (9k - 1)/3. It's now easy to show what any fraction of this form yields an integer when substituted into the original equation.   

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