With n buckets, how many numbers from 1 to n can I put without two buckets having the same pair of numbers. Given n buckets, how many numbers from 1 to n can I throw into the buckets without having two buckets that share the same pair of numbers? The same number can be thrown into different buckets but not twice in the same bucket.
Thanks.

Suppose that there are $k_i$ numbers in each bucket. Let $ K = \sum k_i$. Let's count the number of pairs that are in each bucket. There are $ \sum {k_i \choose 2} $ pairs. By convexity, we have
$$ \sum{k_i \choose 2 } \geq n { \frac{K}{n} \choose 2 } $$
From your condition that no pair of numbers appear in different buckets, this number is bounded above by ${n \choose 2} $.
Hence, we have $\frac{ n(n-1) } { 2} \geq n \frac{ (K/n) ( K/n-1) } { 2}  \Rightarrow  n^2(n-1) \geq K(K-n)$, This gives us
$ K \leq n + \sqrt{ 4n^3 - 3n^2 }.$
As to whether we can achieve (close to) equality, that will depend on the value of $n$.
