3 Fixed typo.

I'll expand slightly on Dan's suggestion.

Any such pair of sequences $(\{a_{i}\}, \{b_{j}\})$ can be obtained from the pair of sequences $(\{i\}, \{n+j\})$ {2n+1-j\})$by finitely many iterations of the following operation. For some$i$and some$j$with$a_{i} = k, b_{j} = k+1$, set$a_{i} = k+1, b_{j} = k$. Now just check that the claim holds for the pair of sequences$(\{i\}, \{n+j\})$, {2n+1-j\})$, and that the operation I described leaves the sum invariant.

2 Fixed TeX.

I'll expand slightly on Dan's suggestion.

Any such pair of sequences $({a_{i}}, {b_{j}})$ (\{a_{i}\}, \{b_{j}\})$can be obtained from the pair of sequences$a_{i} = i, b_{j} = n+j$(\{i\}, \{n+j\})$ by finitely many iterations of the following operation.

For some $i$ and some $j$ with $a_{i} = k, b_{j} = k+1$, set $a_{i} = k+1, b_{j} = k$.

Now just check that the claim holds for the pair of sequences $a_{i} = i, b_{j} = n+j$, (\{i\}, \{n+j\})$, and that the operation I described leaves the sum invariant. I guess I failed to use the pigeonhole principle, but this is the approach that makes the most sense to me. 1 I'll expand slightly on Dan's suggestion. Any such pair of sequences$({a_{i}}, {b_{j}})$can be obtained from the pair of sequences$a_{i} = i, b_{j} = n+j$by finitely many iterations of the following operation. For some$i$and some$j$with$a_{i} = k, b_{j} = k+1$, set$a_{i} = k+1, b_{j} = k$. Now just check that the claim holds for the pair of sequences$a_{i} = i, b_{j} = n+j\$, and that the operation I described leaves the sum invariant.

I guess I failed to use the pigeonhole principle, but this is the approach that makes the most sense to me.