Arthur Engel Problem Solving Strategies infinite descent proof contradiction Ch-14 Q11 The question goes as follows:

$2n+1$ integral weights are given, where $n \geq 1$. If we remove any of the weights, the remaining $2n$ weights can be split into two heaps of equal weights. Prove that all weights are equal.

I was discussing the book with my teacher and she told me she found a contradiction to the statement. Basically, take $2n$ weights of weight $1$ (say $blue$ weights) and one weight of weight $2n-1$ (say $red$ weight). Here, if we remove one blue weight, then we can split with one red weight one side and the rest $2n-1$ blue weights on the other. If we remove the red weight, then we can split the rest blue weights into two partitions of size $n$. This partition satisfies the given condition and all the weights are not equal here.
So are we wrong here? How so? Because I find it hard to digest Arthur Engel being so wrong because in the solution section, he says this can be extended to rational and irrational weights as well.
Also, if we aren't, where is the step in Arthur Engel's proof which overlooks our case?
I'll put the original solution here as well, exactly as it is given in PSS:
Let $w_1,...,w_{2n+1}$ be the integral weights. Since any $2n$ of the weights balance, the sum of any $2n$ of the weights is even. This implies they are all of the same parity. If they are even, we set $w_i \leftarrow \frac{w_i}{2}$, and if they are odd, we set $w_i \leftarrow \frac{w_i-1}{2}$. In each case we get a new set of weights with the same balancing property. Applying this reduction repeatedly, we see that the $w_i$ are congruent mod $2^k$ for all $k$. This implies that all $w_i$ are equal.
Generalize the result to rational weights, which is easy, and to irrational weights, which is much more difficult.
 A: As I mention in the comments, I believe that in the problem we should have restricted that the two heaps we split into have equal size. To justify this, I will explain how the solution fails without this solution, but works with it.
The crucial step is the reduction $w_i\leftarrow\frac{w_i-1}{2}$. Engel claims that "[...] we get a new set of weights with the same balancing property", but this is not necessarily true - if we apply this to the example your teacher gave (say with $n=2$), $1,1,1,1,3$, then this has the balancing property as you mention, but after reduction it doesn't - we get $0,0,0,0,1$, and after removing a zero we can't split it into two heaps of equal weight, since one would have to have weight $0$, other weight $1$.
So let's try to see why an "obvious" argument doesn't give us that the balancing property is preserved. After we remove one weight, say $w_1$, we can split the other ones into two heaps, say $w_2,\dots, w_{k+1}$ and $w_{k+2},\dots,w_{2n+1}$ of size $k$ and $2n-k$, so that $w_2+\dots+w_{k+1}=w_{k+2}+\dots+w_{2n+1}=W$. We would expect that this balancing would lead to a balancing in a new set of weights. Let's see if it does:
$$\frac{w_2-1}{2}+\dots+\frac{w_{k+1}-1}{2}=\frac{w_2+\dots+w_{k+1}}{2}-\frac{k}{2}=\frac{W}{2}-\frac{k}{2},\\
\frac{w_{k+2}-1}{2}+\dots+\frac{w_{2n+1}-1}{2}=\frac{w_{k+2}+\dots+w_{2 +1}}{2}-\frac{2n-k}{2}=\frac{W}{2}-\frac{2n-k}{2}.$$
These are equal only when $k=2n-k$, i.e. when the heaps have equal sizes. So in general, we have no guarantee the new weights have the balancing property, unless we add the restriction in the problem that the heaps have to have equal sizes.
