1. Write the series in a clearer form

The given sum is

$$\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\cdots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}.$$

This can be written as

$$\sum_{k=1}^{9} \frac{1}{(20k-a)(20(k+1)-a)}=\frac{1}{256}.$$

There are $9$ terms corresponding to pairs $(20,40),(40,60),\dots,(180,200)$.

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2. Use partial fractions on the general term

Observe that

$$\frac{1}{(x-a)(x+20-a)} =\frac{1}{20}\left(\frac{1}{x-a}-\frac{1}{x+20-a}\right).$$

So,

$$\frac{1}{(20k-a)(20(k+1)-a)} =\frac{1}{20}\left(\frac{1}{20k-a}-\frac{1}{20(k+1)-a}\right).$$

Hence the sum telescopes:

$$S=\frac{1}{20}\sum_{k=1}^{9}\left(\frac{1}{20k-a}-\frac{1}{20(k+1)-a}\right).$$

Therefore,

$$S=\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right).$$

Given $S=\frac{1}{256}$, we get

$$\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{256}.$$

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3. Simplify the equation

Inside the bracket,

$$\frac{1}{20-a}-\frac{1}{200-a} =\frac{(200-a)-(20-a)}{(20-a)(200-a)} =\frac{180}{(20-a)(200-a)}.$$

So,

$$\frac{1}{20}\cdot \frac{180}{(20-a)(200-a)}=\frac{1}{256}.$$

Thus,

$$\frac{9}{(20-a)(200-a)}=\frac{1}{256}.$$

Cross-multiplying,

$$(20-a)(200-a)=9\cdot 256=2304.$$

Expand:

$$a^2-220a+4000=2304.$$

So,

$$a^2-220a+1696=0.$$

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4. Solve the quadratic equation

$$a=\frac{220\pm\sqrt{220^2-4\cdot 1696}}{2}.$$

Compute the discriminant:

$$220^2=48400,$$ $$4\cdot 1696=6784,$$ $$\Delta=48400-6784=41616.$$

Now,

$$\sqrt{41616}=204$$

since

$$204^2=41616.$$

Therefore,

$$a=\frac{220\pm 204}{2}.$$

So the two values are

$$a=\frac{424}{2}=212,$$

and

$$a=\frac{16}{2}=8.$$

Hence the maximum value of $a$ is

$$\boxed{212}.$$

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5. Check options

• A: $198$ ❌
• B: $202$ ❌
• C: $212$ ✅
• D: $218$ ❌

So the correct option is C.