In the first condition there is,

1. for each k=1,â€¦,nâˆ’1 either x_k>0 or x_(k+1)=0

OK

I see. I badly misread the original problem. I apologize.

$Given:\ k,\ n \in \mathbb Z,\ n > 1,\ 1 \le k \le n,$

$x_k \in \mathbb Z,\ x_k \ge 0,\ and\ \displaystyle \sum_{i=1}^n ix_i = n.$

$Prove:\ \text{The number of solutions such that }k < n \implies either\ x_k > 1\ or\ x_{k+1} = 0$

$\text{equals the number of solutions such that }k \le n \implies either\ x_k = 0\ or\ x_k = 1.$

Do I have the problem correctly?

I still like to look at a few specific cases because that frequently gives a clue on how to solve the general problem. Let's start with n = 2.

$x_k \not > 2.$

$\therefore x_k = 0,\ 1,\ or\ 2.$

$\text{(0, 0), (0, 2), (1, 0), (1, 1), (1, 2), (2, 1), (2, 2) are not solutions.}$

$\text{(0, 1) and (2, 0) are solutions.}$

The first condition but not the second applies to the second solution so that number is 1.

The second condition but not the first applies to the first solution so that number is also 1. The numbers are equal.

I would now try at least n = 3 and maybe even n = 4 and n = 5. I suspect that exploration has a high probability of showing a path to a general proof. For example, it is now fairly obvious that

$\text{(0, ... 1) is always a solution that satisfies the second condition and}$

$\text{(n, 0 ...) is always a solution that satisfies the first condition.}$