Let $0 \leq d \leq K \leq D_L \leq 100$, with $K$ and $D_L$ fixed.
How can I see that $$\dfrac{\displaystyle\binom{100-D_L}{K-d}\binom{D_L}{d}}{\displaystyle\binom{100}{K}}$$ is maximized when $d$ is minimized?
Unfortunately, all I've done is written out the definition $$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$$ for the three above binomial coefficients, and I'm stuck there.
You ask for the maximum value of your expression with respect to $d$.
Let us name your expression
$$ B(d) = \dfrac{\displaystyle\binom{100-D_L}{K-d}\binom{D_L}{d}}{\displaystyle\binom{100}{K}} $$
Then we have the quotient, after evaluating the factorials: $$ Q(d) = \frac{B(d)}{B(d+1)} = \frac{(d+1)(101-D_L - K + d)}{(D_L - d)(K-d)} $$
The claim is that the maximum of $B(d)$ will be obtained at some value of $d$ which we may call $d_m$. This is equivalent to (1) having that $Q(d) > 1$ for all $d \ge d_m$ and (2) having that $Q(d-1) < 1$ for all $d\le d_m$
Starting with (1), $Q(d) > 1$ is equivalent to $$ (d+1)(101-D_L - K + d) > (D_L - d)(K-d) $$ or $$ d > \frac{-101 +K + D_L (K+1)}{102} $$
With the same argument, for (2), $Q(d-1) < 1$ is equivalent to
$$ d < 1 + \frac{-101 +K + D_L (K+1)}{102} $$
So both conditions give the solution, rounding to the next highest integer, $$ d_m = \lceil{\frac{K D_L + (-101 + K+D_L)}{102}} \rceil $$
Some discussion:
For large $K$ and $D_L$, the approximate value is $K D_L / 100$, as expected (see also the comment by polfosol).
For not so large numbers of $K$ and $D_L$, the linear term $(-101 + K+D_L)/102$ induces small shifts.
We will not always obtain positive values for $d_m$, since $d_m$ becomes zero for $-101 +K + D_L (K+1) <0$ or
$$ D_L < \frac{101 - K}{K+1} $$