Assume that ${\rm char}=0$. If $X$ is Fano and $\mathcal L$ is an ample invertible sheaf contained, but not equal to $\omega_X^{-1}$, then $H^i(X,\mathcal L^{-1})=0$ for all $i$. This is Kodaira vanishing for $i\neq \dim X$ and for $n=\dim X$, $H^n(X,\mathcal L^{-1})$ is dual to $H^0(X, \mathcal L\otimes \omega_X)$ which is $0$ by the choice of $\mathcal L$. 

I think both of your examples are covered by this.

In general, if $\mathcal L$ is contained in $\omega_X$, then $H^n(X,\mathcal L)\neq 0$, so for a K3 (or more generally if $K_X=0$) negative ample invertible sheaves will not work. It is also relatively easy to see that if ${\rm Pic}\\, X = \mathbb Z$, then there are no such invertible sheaves. 

**Edit**: 
On the other hand, it is relatively easy to find invertible sheaves with this property on K3's with higher Picard numbers. For instance, let $X$ be a K3 and $L_1, L_2$ two disjoint $(-2)$-curves. This requires the Picard number to be at least $3$, but then it is quite common. For instance, every Kummer has plenty of disjoint $(-2)$-curves. So, let $\mathcal L=\mathcal O_X(L_1-L_2)$. Then $\mathcal L$ has no cohomology for any $i$. 

This can be seen the following way: First notice that the sheaf $\mathcal O_X(L_1)$ has a $1$-dimensional $H^0$ since $L_1$ has negative self-intersection, $H^2=0$ since its dual has no $H^0$, and $H^1=0$ by Riemann-Roch. Then in the short exact sequence
$$
0 \to \mathcal L \to \mathcal O_X(L_1) \to \mathcal O_{L_2} \to 0 
$$
the two sheaves on the right have the exact same cohomology groups and clearly the induced map on $H^0$ is an isomorphism, so the sheaf on the right cannot have any non-zero cohomology. 

Based on this one can make up lots of invertible sheaves on K3 surfaces with this property.