The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$.
This question is guessable but let's prove our answer
\[a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}\]

\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\]

\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\]

lets put $n+1$ into $n$ now

\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\]

and set them equal now

\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n\]

\[a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n\]

let's rewrite it

\[(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n\]

Let's make it look nice and let $b_n=a_{n + 2}-a_n$

\[(b_{n+1})(a_{n + 2}+1)= b_n\]

Since $b_n$ and $b_{n+1}$ are integers, we can see $b_n$ is divisible by $b_{n+1}$

But we can't have an infinite sequence of proper factors, unless $b_n=0$

Thus, $a_{n + 2}-a_n=0$

\[a_{n + 2}=a_n\]

So now, we know $a_3=a_1$

\[a_{3} = \frac {a_1 + 2009} {1 + a_{2}}\]

\[a_{1} = \frac {a_1 + 2009} {1 + a_{2}}\]

\[a_{1}+a_{1}a_{2} = a_1 + 2009\]

\[a_{1}a_{2} = 2009\]

To minimize $a_{1}+a_{2}$, we need $41$ and $49$

Thus, our answer $= 41+49=\boxed {090}$