April wrote: (a) Do there exist 2009 distinct positive integers such that their sum is divisible by each of the given numbers? Suppose it exists $ n$ distinct positive integers $ a_i$ such that $ \sum\frac{1}{a_i}=1$, then let $ b_k=\frac{\prod a_i}{a_k}$ : $ \{b_k\}$ are $ n$ distinct positive integers whose sum is $ S=\sum b_k=\prod a_i\sum\frac{1}{a_k}=\prod a_i$ and each $ b_k$ divides $ S$. In order to end the demonstration, it's easy to show with induction that it is always possible, when $ n>2$, to find $ n$ distinct positive integers $ a_i$ such that $ \sum\frac{1}{a_i}=1$ : It's true for $ n=3$ : $ 1=\frac 12+\frac 13+\frac 16$ If it's true for $ n$, just replace $ \frac 1a$, for the greatest $ a$ with $ \frac 1{a+1}+\frac 1{a(a+1)}$ and so it's true for $ n+1$ Hence the result.

b) The answer is no . We can solve problem with m numbers instead of 2009 .Suppose the existence of such m numbers .We can assume that $ x_1<x_2<...<x_m$ .Let $ s=x_1+...+x_m$ ,by the condition we have $ x_n+x_j|s,\forall s=1,..,m-1$ . We also have : $ m>\frac{s}{x_m+x_1}>\frac{s}{x_m+x_2}>...>\frac{s}{x_m+x_{m-1}}>1$ ,but it gives contradiction to the fact that all number are integers .

TTsphn wrote: b) The answer is no . We can solve problem with m numbers instead of 2009 .Suppose the existence of such m numbers .We can assume that $ x_1 < x_2 < ... < x_m$ .Let $ s = x_1 + ... + x_m$ ,by the condition we have $ x_n + x_j|s,\forall s = 1,..,m - 1$ . We also have : $ m > \frac {s}{x_m + x_1} > \frac {s}{x_m + x_2} > ... > \frac {s}{x_m + x_{m - 1}} > 1$ ,but it gives contradiction to the fact that all number are integers . Very nice !! Just a very little precision, you need $ m>2$ in order to have $ \frac {s}{x_m + x_{m - 1}} > 1$

See topic $ \textbf{hard about set}$ by mathnice1 at http://www.mathlinks.ro/viewtopic.php?t=295676.

(a) It is a well known result that for any $n$, one can choose $1/a_1+...+1/a_n=1$ distinct integers. This is called "Egyptian fractions". Now let the numbers be $(\prod a_i)/a_k$ for $k=1,...,n$. (b) No. Take the 2008 greatest sums $s_1,...,s_{2008}$ and let $S$ be total sum. then $S/s_i$ are 2008 distinct integers strictly between 1 and 2008, impossible!