It's not too difficult! Only attend that at most, there can be $3$ odd numbers and we must only choose an odd number and an even number as $a, b$. So ...

Can you be more specific please because I dont understand it completely ... ?

As EQSon said, there cannot be more than three odds on the blackbord. In fact, if we operate on two odds we get one odd and one even (number of odds is decreased), if we operate on an odd and an even we get an odd and even (number of odds is the same), if we operate on two evens we get two evens (number of odds is the same). Hence, the initial number of odds cannot be increased. Moreover we cannot operate on two odds if we want to reach the required situation. So we can get the first $2009$ only by an $a+b$. So the other (even) number $ab$ left from the operation is $>2009$ (it's easy to exclude the case when one of $a$ and $b$ equals $1$). Therefore we cannot use $ab$ to get the other $2009$'s. Eventually we will be left with two $2009$, two evens that we cannot use to get the third $2009$ and an odd.

You're completely right, dear motal! Thank you

The integers $1$, $2$, $3$, $4$, and $5$ are written on a blackboard. It is allowed to wipe out two integers $a$ and $b$ and replace them with $a + b$ and $ab$. Is it possible, by repeating this procedure, to reach a situation where three of the five integers on the blackboard are $2009$? Mod: Editing the first post with this. For posterity, the first post was originally: Quote: There are five numbers ($1, 2, 3, 4, 5$) on a blackboard. Each step, we can erase two of numbers (like $a, b$) and instead of those two numbers, we write $ab$ and $a+b$. Is it possible to have exactly three $2009$ on the board?

I just posted the problem statement as it was in the contest as well for the Resources section. For the solution scroll up.