We have a house to sell. Each day an offer of $X_n$ comes for the house. Each offer costs an amount $k$ to observe. You may think of $k$ as advertisement costs. When you receive an offer you must decide whether to accept the offer or keep waiting for a better one. In case you accept the offer, your actual profit will be $X_n - nk$.
Now we wish to maximize the amount of money we get paid by choosing a proper stopping rule.
P.S. The problem can be found under the title of "job search problem" in economics literature. How many searches should an unemployed person looking for a job undertake, considering each search costs him.
The answer to your question depends on what you know about the distribution of the $X_i$. Also, the answer changes whether you are allowed to go back to previously encountered offers. From your question it seems you assume we cannot go back.
1) Suppose you don't know anything about the distribution, not even its support $[\underline{x},\overline{x}]$. Then there may not be a solution. After all, search cost $k$ may be larger than $\overline{x}$, so no matter which offer you get you always make losses by searching, but you only learn of this after searching!
2) Suppose you know the distribution $F(x)$ of the offers, and offers are iid. Now you can compute an optimal stopping rule, because it is possible to trade-off the search costs you incur by searching further with the (expected) benefit. The optimal stopping rule is a threshold rule $\alpha$ which satisfies $$\int_\alpha^\infty (x-\alpha) dF(x)=c,$$ as shown in Sakaguchi 1961, Dynamic programming of some sequential sampling design, Journal of Mathematical Analysis and Applications. An analogue result if recall is allowed is given in Lippman&McCall 1976, The Economics of Job Search Part I, Economic Inquiry.
3) If I recall correctly, there were some results in the literature which showed that the optimal stopping rule for some classes of the problem even if you don't know the distribution $F$ is also a threshold strategy.