Harvard-MIT Mathematics Tournament

Solution:
This is similar to the previous solution. If $d$ is a divisor of $i$, then so is $i/d$, and $(i/d)/i = 1/d$. Summing over all $d$, we see that $\sigma(i)/i$ is the sum of the reciprocals of the divisors of $i$, for each positive integer $i$. So, summing over all $i$ from $1$ to $n$, we get the value $1/d$ appearing $\lfloor n/d \rfloor$ times, once for each multiple of $d$ that is at most $n$. In particular, the sum is
$$\frac{1}{1}\left\lfloor\frac{n}{1}\right\rfloor+\frac{1}{2}\left\lfloor\frac{n}{2}\right\rfloor+\frac{1}{3}\left\lfloor\frac{n}{3}\right\rfloor+\cdots+\frac{1}{n}\left\lfloor\frac{n}{n}\right\rfloor < \frac{n}{1^{2}}+\frac{n}{2^{2}}+\cdots+\frac{n}{n^{2}}.$$
So now all we need is $1/1^{2} + 1/2^{2} + \cdots + 1/n^{2} < 2$. This can be obtained from the classic formula $1/1^{2} + 1/2^{2} + \cdots = \pi^{2}/6$, or from the more elementary estimate
$$\begin{aligned} 1/2^{2} + 1/3^{2} + \cdots + 1/n^{2} &< 1/(1 \cdot 2) + 1/(2 \cdot 3) + \cdots + 1/((n-1) \cdot n) \\ &= (1/1 - 1/2) + (1/2 - 1/3) + \cdots + (1/(n-1) - 1/n) \\ &= 1 - 1/n \\ &< 1. \end{aligned}$$