NP-Completeness of modified decision problems
It depends on the problem. For, say, clique or independent set, the problems become trivial if you change the direction of the inequality: $\emptyset$ is always a clique/independent set of size&nbsp;$\leq k$.

On the other hand, the travelling salesman problem remains NP-complete if you ask for a tour of length at least&nbsp;$d$ instead of at most&nbsp;$d$.  To see this, suppose that you want a tour of graph&nbsp;$G$ of length at most&nbsp;$d$. Let $m$ be the length of the longest edge in&nbsp;$G$, and let $G'$ be the graph that's identical to&nbsp;$G$ except that, if an edge has length&nbsp;$\ell$ in&nbsp;$G$, then it has length $m-\ell$ in&nbsp;$G'$.  Now, any tour of length&nbsp;$t$ in&nbsp;$G$ corresponds to a tour of length&nbsp;$nm-t$ in&nbsp;$G'$ (where $n$&nbsp;is the number of vertices), so $G$&nbsp;has a tour of length at most&nbsp;$d$ if, and only if, $G'$&nbsp;has a tour of length at least $nm-d$.