Exactly, your language $L_k$ can be accepted by a minimal DFA of $k$ states because it needs to "save" the remainder in the state itself. A formal proof would require the Myhill-Nerode theorem and counting the of the remainders induced equivalence classes.
If we consider the language $L_i \cap L_j$ for $i > j > 1$, it is not generally true that the minimal automaton needs $i \cdot j$ states because if you set $i = 6$ and $j = 4$ you'll get the language $$L = \{w \mid |w| \text{ is divisible by } 4 \text{ and } 6\}$$
It is a known fact that if an integer is divisible by $i$ and $j$ it also is divisible by their least common multiple (proof here: https://proofwiki.org/wiki/LCM_Divides_Common_Multiple), this leads to:
$$L= \{w \mid |w| \text{ is divisible by } 12\}$$
Using the upper arguments it follows that a DFA accepting $L$ needs $12$ states and using your conjecture the automaton would need $24$ and is therefore not minimal.
Now back to your problem with $i = 7$ and $j = 2$. Using the fact that common multiples of $7$ and $2$ are divisible by $\operatorname{lcm}(2,7)=14$, it leads to the conclusion that: $$L_i \cap L_j = \{w \mid |w| \text{ is divisible by } 14\}$$
Therefore the conclusion using the Myhill-Nerode theorem is that an automaton accepting $L_i \cap L_j$ requires at least $14$ states.
