Solution:

Since the lcm of $x_{i}$ and $x_{j}$ is greater than $N$, among the numbers $1,2, \ldots, N$ there are no two that are multiples of both $x_{i}$ and $x_{j}$.
Among the multiples of the natural number $x$ there are two such that $k x \leq N < (k+1) x$, from which it follows that $k \leq \frac{N}{x} < k+1$. The number of multiples of $x$ that are less than $N$ is accordingly the integer part of $\frac{N}{x}$, i.e. equal to $\left[\frac{N}{x}\right]$.

For the numbers $x_{1}, x_{2}, \ldots, x_{n}$ it therefore holds that $\left[\frac{N}{x_{1}}\right]+\left[\frac{N}{x_{2}}\right]+\cdots+\left[\frac{N}{x_{n}}\right]<N$.

On the other hand, $\left[\frac{N}{x_{i}}\right]>\frac{N}{x_{i}}-1$ and hence $\frac{N}{x_{1}}+\frac{N}{x_{2}}+\cdots+\frac{N}{x_{n}}-n<N$.

But since $n<N$, it follows that:
$\frac{N}{x_{1}}+\frac{N}{x_{2}}+\cdots+\frac{N}{x_{n}}<2N$, which finally leads to $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}<2$.