Let $X_{k}$ denote the number of boxes that are to be bought to come in possession of $k+1$ tokens, counting from the moment that one is in possession of exactly $k$ different tokens.
Then $X=1+X_{1}+\cdots+X_{n-1}$ boxes must be bought.
Here $X_{k}$ has geometric distribution with parameter $p_{k}=1-\frac{k}{n}$ so that $\mathbb{E}X_{k}=\frac{1}{p_{k}}=\frac{n}{n-k}$.
We find $\mathbb{E}X=\sum_{k=0}^{n-1}\frac{n}{n-k}=n\sum_{k=1}^{n}\frac{1}{k}$
The expected amount of money to be paid is $nx\sum_{k=1}^{n}\frac{1}{k}$.