I have the following maximization problem, which I believe should have been studied before.
Let $n=3$ and fix $t$. I want to maximize $-\Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \Big)+\Big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \Big)-\frac{1}{a+b+c}$ where the maximization is over $a,b,c>0$ tuples of integers satisfying $a+b+c=t$.
Note that this is a baby version of the case for general $n$. So for general $n$, I want to maximize a sum (similar to the one above with alternating signs) with $n$ variables $a_1,\ldots,a_n$ (satisfying $\sum_i a_i=t$).
Edit: After considering that $a, b, c$ are positive integers, I believe you can achieve the maximum value by keeping $a, b, c$ as close to as $t/3$, i.e the average of them. Also, if there are $n$ variables, then you should set them around $t/n$. I'd say, give each of them $[|t/n|]$ and give some of them 1 more as many times as the remainder. Like for $t=25$ and $n=4$, go for $6,6,6,7$.
But this is just what I feel after a very brief observation; I have no proof for it. If what I say is correct, then it is still a hard work to calculate that maximum value, because that almost-equal distribution of $t$ to $n$ variables will not be fully equal because of the condition that variables are integers and I guess for every remainder of $t/n$ you should make a calculation to find out that maximum number. But some good news, if what I say is correct, then you can make a computer compute it for you.
That expression is maximally infinity. Just take $b=c=t/2$ and choose $a=-0$ (a approaches to zero from the left side), and $1/a$ will become negative infinity and the other fractions will remain finite and notice that there is a minus behind the parantheses where $1/a$ takes place which will make it positive infinity. Similarly if you choose $a=+0$, then you will get minus infinity as the minimal value.
End of Edit 1
Edit 2:
If t has a remainder of 0 in divison by 3, then let $t=3x$, then set $a=b=c=x$, according to my approach maximum should be $-3/(2x)$
If t has a remainder of 1 in division by 3, then let $t=3x+1$, then set $a=b=x$ and $c=x+1$, according to my approach maximum should be $-1/(2x(x+1))$
If t has a remainder of 2 in division by 3, then let $t=3x+2$, then set $a=x$, $b=c=x+1$, according to my approach maximum should be $-1/(2x(x+1)(2x+1))$
But again for a generalized solution for $n$ based on the the remainder of $t$ divided by $n$, you should try harder, there must be a pattern.
End of Edit 2
For your expression to have maximally finite limit, you should remove the first minus sign. Then the solution will be as follows:
For n=3: Use the inequality that the arithmetic mean is bigger than or equal to harmonic mean (they are equal only when all elements are the same).
$$(a+b+c)/3 > 3/(1/a + 1/b + 1/c)$$ $$t/3 > 3/(1/a + 1/b + 1/c)$$ $$t/9 > 1/(1/a + 1/b + 1/c)$$
$$((a+b)+(b+c)+(c+a))/3 > 3/(1/(a+b) + 1/(b+c) + 1/(c+a))$$ $$2t/3 > 3/(1/(a+b) + 1/(b+c) + 1/(c+a))$$ $$2t/9 > 1/(1/(a+b) + 1/(b+c) + 1/(c+a))$$
$$1/(a+b+c) = 1/t$$
So,
$$t/9 + 2t/9 - 1/t > 1/(1/a + 1/b + 1/c) + 1/(1/(a+b) + 1/(b+c) + 1/(c+a)) - 1/(a+b+c)$$
That means $1/(1/a + 1/b + 1/c) + 1/(1/(a+b) + 1/(b+c) + 1/(c+a)) - 1/(a+b+c)$ can maximally become $t/3 - 1/t$
For $n=n$ the answer is similar: $3t/n^2 - 1/t$
What allowed maximally infinite value for your expression is that, the same inequality worked in opposite ways, because one of your fractions had the opposite sign of the other. In such cases, appropriately chosen values usually allow you to reach infinity. As an exercize, choose $a=t$ and $b=+/-0, c=+/-0$ and analyze.