Is the sum of all irrational numbers between any two integer constants rational?
I think it should be, because every irrational number should have another irrational with which it would sum to a rational number (e.g. 3.141... sums with 3.858..., and 2.718... with 2.281...).
Why is this or isn't this a reasonable conclusion?
Addition is only done between pairs of numbers: you can add $a$ and $b$ to obtain $a+b$. We can extend this to sums of more than two numbers because addition is associative and commutative: when adding $a,b, $ and $c$ we don't have to worry about whether we want $(a+b)+c$ or $a+(b+c)$ or $(b+c) + a$ or whatever because the answer is always the same.
In some cases we can even extend this notion of summation to certain infinite sums. For example, there is a well-defined sense in which $$1+\frac12+\frac14+\frac18+\cdots = 2$$ because we can consider the sequence $1, 1+\frac12, 1+\frac12+\frac14,\ldots$ and show that it approaches 2 from below.
However, none of this is any help in adding up all the irrational numbers, even those between two integers, because the irrational numbers have a property called uncountability: however you consider an infinite sum $i_1, i_1+i_2, i_2+i_2+i_3, \ldots$ of irrational numbers, you must always omit some irrational numbers from your list $i_1, i_2, i_3, \ldots$. In fact you must omit most irrational numbers. So the idea "sum of all the irrational numbers" simply has no mathematical meaning. The notion of addition does not extend to adding up that many things.
Even for relatively well-behaved sets like the set of all rational numbers (which are not uncountable), there is no clear notion of the sum of all the members of the set. It can be shown that by ordering the rational numbers in the correct order $r_1, r_2, r_3, \ldots$, one can make the sums $r_1, r_1+r_2, r_1+r_2+r_3$ approach any desired target, or diverge to infinity.
If one considers just the set of rationals between 0 and 1, as you suggested, the sum diverges to infinity. To see this, suppose the sum of all rational numbers between 0 and 1 added up to some number $S$. Now consider the following sums: $$ \begin{align} S_1 & = \frac12 + \frac14+\frac18 + \cdots &&= 1 \\ S_3 & = \frac35 + \frac3{10}+\frac3{20} + \cdots &&= \frac65 \\ S_5 & = \frac59 + \frac5{18}+\frac5{26} + \cdots &&= \frac {10}9 \\ &\vdots \end{align} $$
All together, $S_1, S_3, S_5, \ldots$ include only some of the rational numbers between 0 and 1, so $S$ must be bigger than $S_1$, it must be bigger than $S_1+S_3$, it must be bigger than $S_1+S_3+S_5$, and so on. But since each of the $S_1, S_3, S_5,\ldots$ is at least 1, this shows that $S$ must be bigger than 1, bigger than 2, bigger than 3, and so on. Since there is no such number $S$, it is not the case that the sum of all rational numbers is a number, and it does not make sense to ask whether that number is rational or irrational, because there is no such number.
So in mathematics as it is normally practiced at present, your question doesn't make sense and has no answer, neither 'yes' nor 'no'.