Solution:

We shall show by induction on $k\left(k \leqslant \frac{2n-1}{3}\right)$ that one can discard $3k$ of the subsets so that the cardinality of the union of the remaining ones is no greater than $n-k$.

For $k=0$ this is trivial. Suppose that $k \geqslant 1$ and that we have discarded $3(k-1)$ subsets so that the union of the remaining ones contains no more than $n-k+1$ elements. Since $2n-1-3(k-1)<2(n-k+1)$, there exists an element $x_{k}$ of the union which lies in at most three of the remaining subsets. Thus we can discard three more subsets so that the union of the remaining $2n-1-3k$ subsets does not contain the element $x_{k}$, which completes the induction step.

The assertion of the problem follows for $k=\left[\frac{n-1}{3}\right]$, since $n-\left[\frac{n-1}{3}\right] \leqslant n-\frac{n-3}{3}=\frac{2}{3}n+1$.