I have show that the number of solutions $\left(\, i,j\,\right)$ of non-negative integers to $i + jk \leq l$ is $$ \left(\,\left\lfloor\, l \over k\,\right\rfloor +1\,\right) {2l + 2 - k\left\lfloor\, l/k\,\right\rfloor \over 2} $$
I thought of this as the solution to how many lattice points are in the triangle with points $(0,0), (l,0), \left(0, \left\lfloor\,\frac{l}{k}\,\right\rfloor\right)$.
If $h = gcd(l, \lfloor\frac{l}{k}\rfloor) -1$ then there are $l + \lfloor\frac{l}{k}\rfloor + 1 +h$ points on the boundary and $\frac{(l-1)(\lfloor\frac{l}{k}\rfloor -1) -h}{2}$ interior points of the triangle and the solution would be two just add these two together.
Doing that however didn't result in what I was supposed to show, so if somebody could point out where I've gone wrong here I'd appreciate it!
The triangle you want consists of all points in the first coordinate below the line $x+ky \leq l$, so you want the vertices to be $(0,0)$, $(l,0)$, and $(0, \frac{l}{k})$.
You could try to get from your answer to the correct solution by counting all lattice points in the triangle with vertices $\left(0, \left\lfloor\,\frac{l}{k}\,\right\rfloor\right), (0,\frac{l}{k}), (l,0)$, besides the $h+1$ points you've already counted on the line from $\left(0, \left\lfloor\,\frac{l}{k}\,\right\rfloor\right)$ to $(l,0)$, and add those to your total.
I'm not sure if there's an easy way to do that or not. Presumably that piece would include a $-h/2$ term.
By writing $l = mk + a$, for some $a$ with $0 \leq a < k$, I was able to verify algebraically that $$ \sum_{i=0}^{l} \left(\left\lfloor\,\frac{i}{k}\,\right\rfloor+1\right) = \left(\,\left\lfloor\, l \over k\,\right\rfloor +1\,\right) {2l + 2 - k\left\lfloor\, l/k\,\right\rfloor \over 2}, $$ so if your geometrical method doesn't yield fruit, you could try that approach.