Let $\frac{x}{y}=1+\frac12 +\frac13 +\cdots +\frac1{96}$ where $\text{gcd}(x,y)=1$. Show that $97\;|\;x$.
I try adding these together, but seems very long boring and don't think it is the right way to solving. Sorry for bad english.
If you group the fractions in pairs with the first pairing to last, second pairing to next-to-last, etc, you get $$1+\frac 1{96}=\frac {97}{96}, \frac 12+\frac 1{95}=\frac {97}{190}...$$ The sums of these pairs all have a numerator of $97$, and because $97$ is prime the common denominator will not have a factor of $97$, so in $\frac xy$, $x$ is a multiple of $97$.
What we need to do is as follows: $$ \begin{equation}\begin{split} 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{96} & = \Big(1+\frac{1}{96}\Big) + \Big(\frac{1}{2}+\frac{1}{95}\Big) + \ldots + \Big(\frac{1}{48} + \frac{1}{49}\Big) \\ & = \frac{97}{96} + \frac{97}{2*95} + \frac{97}{3*94} + \ldots + \frac{97}{48*49} \\ & = 97 \Big( \frac{1}{96} + \frac{1}{2*95} + \frac{1}{3*94} + \ldots + \frac{1}{48*49}\Big) \end{split}\end{equation} $$
Hence, the numerator is a multiple of $97$. To see that the denominator is not a multiple of $97$, note that $97$ is a prime, and the denominator contains natural numbers less than $97$ (in fact, the reduced denominator is a factor of $96$!), hence cannot possibly be a multiple of $97$. Thus, in it's reduced form, the numerator of the expression is a multiple of $97$.
In fact, Wolstenhomme's theorem says that the numerator is a multiple of $97^2 = 9409$!
This is along the lines of DanielV's proposed solution.
We know $97$ is prime. Multiplying both sides by $96!$, it's enough to show that $\sum_{n=1}^{96} \frac{96!}{n}$ (which is a sum of integers) is divisible by $97$, because $96!$ and $97$ are coprime. Let $a_n = \frac{96!}{n}$ for $n\in\{1,\ldots,96\}$. Then we have $n a_n = 96!$, so $a_n \equiv 96! n^{-1} \pmod {97}$, where $n^{-1}$ is the inverse of $n$ in the integers mod $97$. Then $\sum_{n=1}^{96} a_n$ is divisible by $97$ because $$ \sum_{n=1}^{96} a_n \equiv 96! \sum_{n=1}^{96} n^{-1} \color{red}{=} 96! \sum_{n=1}^{96} n = 96! \frac{96 \cdot 97}{2} \equiv 0 \pmod{97}.$$ The red $\color{red}{=}$ holds because the inverse $n \mapsto n^{-1}$ is a permutation of $\{1,\ldots,96\}$.
If you know that 97 is prime, and that every nonzero value has a multiplicative inverse mod a prime value, and that modular inverse is a bijection due to it being an involution, then you get:
$$\sum_{k = 1}^{96} k^{-1} \equiv \sum_{j = 1}^{96} j \pmod {97}$$
And the result follows from arithmetic sum.
More explicitly, letting $p = 97$,
$p \not | y$ because $y | (p - 1)!$ and $p$ is prime and $(x,y)$ are in reduced terms. Therefore:
$xy^{-1} \equiv 0 \pmod p$ iff $x \equiv 0 \pmod p$ implies the numerator of $\frac{x}{y}$ is divisible by $p$.
Since $\frac{x}{y} = \sum_{k=1}^p k^{-1}$, follows that
$$p | x \text{ iff } \sum_{k = 1}^{p-1} k^{-1} \equiv 0 \pmod p$$
Let $S_n \equiv n^{-1} \pmod p$ with $n$ ranging from $1$ to $p-1$. Then $S$ is a bijection, so
$$\sum_{n = 1}^{p-1} S_n \equiv \sum_{n = 1}^{p-1} n \pmod {p}$$
And the result follows from
$$p(p-1)/2 \equiv 0 \pmod p$$