To determine whether there exists a positive irrational number $x$ such that there are at most finitely many positive integers $n$ satisfying the condition that for any integer $1 \leq k \leq n$, $\{kx\} \geq \frac{1}{n+1}$, we proceed as follows:

Assume for contradiction that there exists such an $x$. This would imply that there exists a positive integer $N$ such that for all $n > N$, the inequality $\{nx\} > \frac{1}{n+1}$ holds. However, by Dirichlet's approximation theorem, for any irrational number $x$ and any positive integer $n$, there exists an integer $k$ such that $1 \leq k \leq n$ and $\{kx\} < \frac{1}{n+1}$. This contradicts our assumption.

Therefore, no such positive irrational number $x$ exists.

The answer is: \boxed{\text{No}}.