A four-part trilogy

First note that for any prime $p$ and positive integer $a$ we have that $f(p^{a}) = ap^{a-1}.$

(This can be verified by induction on $a.$ For $a = 1$ we have $f(p) = 1 = ap^{1-1}.$ If the statement is true for some $a \gt 1$ then

$f(p^{a+1}) = f(p*p^{a}) = pf(p^{a}) + p^{a}f(p) = pap^{a-1} + p^{a} = (a + 1)p^{(a + 1) - 1},$

and thus the statement is true for $a + 1$ as well.)

So for the prime decomposition $n = p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}...$ we can show by a similar induction on the number of primes that

$f(n) = n\left(\dfrac{a_{1}}{p_{1}} + \dfrac{a_{2}}{p_{2}} + \dfrac{a_{3}}{p_{3}} + ... \right).$

Now to have $f(n) = n$ we require that this last bracketed expression, (call it $S_{n}$ ), be equal to $1.$ Since all the denominators are prime, this can only happen if one of the terms $\frac{a_{k}}{p_{k}} = 1$ and all the others equal $0.$ Thus we can only have $f(n) = n$ when $n = p^{p}$ for some prime $p.$ This gives us the value

$A = 2^{2} + 3^{3} + 5^{5} + 7^{7} = 826699.$

Next, $S_{n}$ is equivalent to $\frac{f(n)}{n}.$ We can maximize this by maximizing the exponents and minimizing the prime factors under the condition that $n \le 1000.$ we have $2^{9} = 512$ giving us $\frac{f(n)}{n} = \frac{9}{2}.$ Next down the list is $2^{8}*3$ giving us $\frac{f(n)}{n} = 4 + \frac{1}{3} \lt \frac{9}{2}.$ As we "power down" $2$ we can increase the powers of other prime factors, but not enough to compensate for the lessening power of $2,$ leading us to conclude that $B = \dfrac{9}{2}.$

Now for $C,$ we need to maximize the product of $n$ and $S_{n}.$ We found that $S_{n}$ is maximized for $n = 512,$ but for this value $nS_{n} = 2304.$ We can do better by increasing $n$ while sacrificing $S_{n}$ to a small extent. The next on the list is $n = 2^{8}*3,$ for which $f(n) = 768(4 + \frac{1}{3}) = 3328.$ We can now rule out any $S_{n}$ less than $3 + \frac{1}{3}.$ Looking for such values for larger $n$ we next look at $n = 2^{7}*7,$ for which $f(n) = 3264.$ Next we have $2^{6}*3*5 = 960,$ for which $f(n) = 3392.$ Any lesser $n$ and/or power of $2$ will necessarily yield a smaller $f(n)$ than this, so $C = 3392.$

Finally, since $1000 = 2^{3}5^{3}$ we have $f(1000) = 1000(\frac{3}{2} + \frac{3}{5}) = 2100,$ and since $42 = 2*3*7$ we have $f(42) = 42(\frac{1}{2} + \frac{1}{3} + \frac{1}{7}) = 41,$ making $D = 2141.$

(The "four-part trilogy" in the title is in reference to the $42$ component of this question, in case anyone was wondering.)

Thus $A + 2B + C + D = 826699 + 9 + 3392 + 2141 = \boxed{832241}.$