Why is $\sum_{i=1}^n a$ always irrational if $n>0$ and $a$ is irrational?

It is trivially so. If you sum the irrational number $x$ $n$ times ($n$ being an integer "of course"), you end up with $nx$.

If $nx=\frac ab$, with $(a,b)$ integers, then $x=\frac{a}{nb}$, thus is rational, which is not true...

$\sum k_ia = a(\sum k_i)$ and $\sum k_i$ is rational.

And if $\sum k_i \ne 0$ then a (non-zero) rational times an irrational is irrational.

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I guess it's somewhat important to point out that if the the sum is $0$ such as $a + 2a - 3a$ or $k_1a + k_2 a + ..... + k_n a$ where $k_1 + k_2 + .... + k_n = 0$, then the statement is trivially false.

But that is the only exception.