Calculating concentration of an ionic compound
Keep it as simple as possible :)

If both initial solutions are not saturated and all the $\ce{NaCl}$ is dissolved, this will not change upon mixing.

If in doubt with the result of a calculation, have a look at the units!

Concentrations are given as the quotient of moles and volume, $c = \frac{n}{V}$, the unit apparently is $\mathrm{mol \cdot l^{-1}}$. The absolute amount of $\ce{NaCl}$ in the first solution is therefore given as $n = c\cdot V$. (Note that this is not the n in your question ;).

The new $\ce{NaCl}$ concentration is the sum of the moles, divided by the new total volume:
$c = \frac{(k\cdot n)+(l\cdot m)}{m+n}$

Unless given further information, it is safe to assume that $\ce{NaCl}$ still is fully dissociated after mixing.