Well no. I let n be the "number of leaves per tree".

That is, n is the number of leaves on any given tree, not the number of leaves total.

If we let the number of leaves be n on the other hand we do have other issues.

If we let n > m we have a major problem. I can actually disprove the statement that there must be two Red Maples with the same number of leaves.

We simply have m trees, and I arrange them like so.

The first tree has 0.

The second tree has 1.

The third tree has 2.

The fourth tree has 3.

...

The mth tree has m-1.

Such that n = 1 + 2 + 3 +...+ (m-2) + (m-1)

There's no reason that trees can't have this arrangement of leaves, and the existence of the possibility of the situation automatically disproves the hypothesis.

However if we let n < m, as Kenny said, it's easily provable.

We'll imagine the trees as boxes for now, and the leaves as leaves that we can put in the boxes.

If n < m, we'll just put one leaf in each box until we fill them all. But since n < m we can't fill them all. But we've got a sequence of leaves in each box that looks like this:

1, 1, 1, 1, 1, 1,..., 1, 0, 0, 0,...

Out to the mth box, or tree.

Now we go to the box containing a leaf to the farthest right. We'll take a leaf from that one and add it to the second box. Now we have

1, 2, 1, 1, 1,..., 0, 0, 0,...

Repeat this process, adding each leaf to the next box down the line so that you get this:

1, 2, 2, 2,..., 0, 0, 0...

Now take a leaf from the box to the farthest right again, and this time add it to the third box and repeat.

Keep doing this process and you'll eventually get:

1, 2, 3, 4, 5,..., 0, 0, 0,...

Here's the thing about this entire process. To start we had m-n empty boxes, and which each iteration we only generated more empty boxes, so therefore more trees with 0 leaves.

Any sequence of distinct numbers you offer me has a tail of at least m-n zeros therefore, and therefore at least two Red Maples with the same number of leaves: zero.

Still, however, due to the lack of information the original problem remains unsolvable without another assumption that wasn't given to you in the problem.