Let $\sigma(n)$ be the sum of all positive divisors of $n$, where $n$ is a positive integer (for instance, $\sigma(6) = 12$ and $\sigma(11) = 12$). We say that $n$ is almost perfect if $\sigma(n) = 2n - 1$ (for instance, 4 is almost perfect since $\sigma(4) = 7$). Let $n \mod k$ be the remainder of the division of $n$ by $k$ and $s(n) = \sum_{1 \le k \le n} n \mod k$ (for instance, $s(6) = 0 + 0 + 0 + 2 + 1 + 0 = 3$ and $s(11) = 0 + 1 + 2 + 3 + 1 + 5 + 4 + 3 + 2 + 1 + 0 = 22$). Prove that $n$ is almost perfect if and only if $s(n) = s(n-1)$.