Iranian Mathematical Olympiad

According to the permutation inequality, we always have:
$$1 \times n + 2 \times (n-1) + \cdots + n \times 1 \le 1 \times s_1 + 2 \times s_2 + \cdots + n \times s_n \le 1 \times 1 + 2 \times 2 + \cdots + n \times n$$
We want to prove that all numbers in this range are covered. First, note that by induction, it is easily proven that:
$$1 \times n + 2 \times (n - 1) + \cdots + n = \binom{n+2}{3}$$
and
$$1 \times 1 + 2 \times 2 + \cdots + n \times n = \binom{n+2}{3} + \binom{n+1}{3}$$
First, let's examine the base case $n = 4$. We need to show that all numbers from 20 to 30 are covered. It is sufficient to consider the permutations:
$$4, 3, 2, 1/3, 4, 2, 1/3, 4, 1, 2/4, 2, 1, 3/2, 3, 4, 1/2, 4, 1, \\ 3/1, 4, 3, 2/1, 3, 4, 2/2, 1, 4, 3/2, 1, 3, 4/1, 2, 3, 4.$$
To prove by induction, we first set $x_{n+1} = (n+1)^2$ for all values of $n$ with:
$$(n+1)^2 = 1 \times 1 + 2 \times 2 + \cdots + n \times n + (n+1) \times (n+1) - (1 \times 1 + 2 \times 2 + \cdots + n \times n)$$
All previous values are obtained by setting $x_{n+1} = 1$:
$$\frac{n(n+1)}{2} = 1 \times (n+1) + 2 \times n + \cdots + (n+1) \times 1 - (1 \times n + 2 \times (n-1) + \cdots + n \times 1)$$
So, to show that all numbers are covered, it is sufficient to show that:
$$\binom{n+2}{3} + \binom{n+3}{3} + \left( \binom{n+2}{3} - \binom{n+2}{3} \right) + \binom{n+1}{3} > \binom{n+2}{3} + \binom{n+3}{3} + \left( \binom{n+1}{3} - \binom{n+2}{3} \right)$$
which is equivalent to:
$$\binom{n+2}{3} + 1 < 2 \times \left( \binom{n+1}{3} + 1 \right)$$
which is obvious.