First, note that since $a_1 + a_2 + \cdots + a_n$ is increasing according to the relation $a_1 + a_2 + \cdots + a_n = \frac{x_n}{n}$, the sequence $x_n$ must also be increasing. We prove by induction that $a_n = n$. First, notice that by the problem's condition, it holds up to $n = 5$. Assume the statement holds for $n$, and we want to prove it for $n+1$. For $i < n$, we have:
$$a_i = \frac{1}{i-1} - \frac{1}{i} = \frac{1}{i(i+1)}$$
Thus, we have:
$$a_{n+1} = 2 - \frac{1}{x_{n+1}} - 2 + \frac{1}{n} = \frac{x_{n+1} - n}{n x_{n+1}}$$
Now if $\gcd(x_n, n) = d$, by Bezout's theorem, we have an interval of length $\frac{d}{n x_{n+1}}$. Because $d|x_{n+1} - n \rightarrow d \le x_{n+1} - n$, by assumption, we should have $d = x_{n+1} - n$. If $d = n k$, we have
$$x_{n+1} = d(k+1) \rightarrow a_{n+1} = \frac{1}{n(k+1)}$$
On the other hand:
$$a_{n+1} \le \frac{1}{n(n-1)} \implies k+1 \ge n-1 \implies k \ge n-2$$
Since $n \ge 5$ and we have $k|n$, we have $k=n$, and consequently $x_{n+1} + 1 = n$. ■