Let $\Lambda :=\{\lambda_1, \dots, \lambda_n\}$ be a set of $n$ distinct real numbers.

For a given $p \in \mathbb N$, consider further the set

$$I_p := \{ \{i_1, i_2, \dots, i_p\} : i_j \in \{1, \dots,n\} \text{ for all } j=1, \dots, p\}.$$

It is not hard to see that $|I_p| = \binom{n+p-1}{p}$. For example, if $n=3$ and $p=2$, then

$$ I_2 = \{ \{1,1\}, \{1,2\}, \{1,3\}, \{2,2\}, \{2,3\}, \{3,3\} \}.$$

Now for a given $p \in \mathbb N$ we write down all $\binom{n+p-1}{p}$ sums $$\lambda_{i_1} + \lambda_{i_2} + \dots + \lambda_{i_p}. $$

Clearly it can happen that "different" sums have the same value, e.g. if

$$\Lambda = \{0,1,2\}$$

and $p=2$, then we get the sums $0, 1, 2, 2, 3, 4$ given by considering the elements of $I_2$.

My question is the following: Is there in general an upper bound on how many "different" sums there are that lead to one particular value?

Another interesting question would be the following: Given a fixed upper bound $N^*$, are there conditions on $\Lambda$ such that for all $p \in \mathbb N$, at most $N^*$ different $p$'th order sums yield the same value. For $N^* = 1$ one gets the condition that $\lambda_1, \dots, \lambda_n$ need to linearly independent over $\mathbb Q$.

upperbound on the number of elements of $I_p$ that can have the same sum. In your example that number is $1$, which is a lower bound. $\endgroup$ – Robert Israel Oct 2 '14 at 18:52