Is there a constraint on the sum of the type-I & type II error probabilities? Is it true that if $H_0$ and $H_a$ are complementary hypotheses of the Binomial trial, i.e., the negation of $H_0$ is $H_a$ then the type-I error $\alpha$ plus type-II error $\beta$ equals 1? Or is that sum always less then 1, or can it be sometimes even greater then 1?
 A: Case 1: the null hypothesis is true. The type II error is 0. The type I error is less than the nominal size of the test unless the test is biased. It can be as high as 1 if the test decision is "reject the null every time".
Case 2: the null hypothesis is false. The type I error is 0. the type II error can be as high as 1 if the test decision is "do not reject the null any time".
To conflate Bayesian and frequentist terminology : you can't speak of the Pr(Type 1 error) without "conditioning" or knowing H_0 is true. A nice bit of frequentist notation is this: $P_{H_0}(\text{Event})$ to refer to probabilities of events or outcomes under the probability model where the null is true, or $P_{\theta = \theta_0}(\text{Event})$ equivalently.
If you want to be crazy and sum together probabilities that don't make sense, you can conceive of two values of $\theta \ne \theta_0$ and $\theta=\theta_0$ for which the Type 1 and Type 2 errors add to more than 1. For instance: 

P(decomposing corpse | dead) + P(looking alright | alive) > 1

Is this surprising or interesting? No. IRL one of those error probabilities will always be 0 and the other is less than or equal to 1 depending on how good or stupid the test is.
A: For an arbitrarily chosen decision rule -- meaning that the decision rule can be anything that you make up just for the heck of it, it doesn't need to be sensible in any sense of the word -- the arithmetic sum of the Type I and Type II error probabilities can be any number in $[0,2]$ as the answer by @Bjorn points out.
Example 1: The observation $X$ always has positive value when $H_0$ is true and always has negative value when $H_a$ is true.  The decision rule is
$$X \begin{array}{c}H_0\\\gtrless\\{H_a}\end{array} 0$$
leading to both the Type I and Type II error probabilities having value $0$ and so their sum is also $0$.
Example 2: As in Example 1, the observation $X$ always has positive value when $H_0$ is true and always has negative value when $H_a$ is true.  But now the decision rule is
$$X \begin{array}{c}  H_a\\\gtrless\\{H_0}\end{array} 0$$
which is exactly bass ackwards from the decision rule in Example 1 (Hey, I said upfront that we are going to consider arbitrary decision rules, not necessarily only sensible ones!!).  Now, since the OP asks in a comment for an explanation of my assertion (in a previous version of this answer) that "both the Type I and Type II error probabilities have value $1$ and so their sum is $2$.", here goes.  

If the null hypothesis is true, then in our model, the observation $X$ is always a positive number. But the decision rule is that whenever the observation has positive value, we are going to reject the null hypothesis. Continuing to remember that the null hypothesis is true, we see that our decision rule tells us to always reject the null hypothesis (when it is true).  So, what is the probability that we reject the null hypothesis when in fact the null hypothesis is true? $100\%$, right? A similar argument applies to the Type II error probability.  If the null hypothesis is actually false, the observation must have negative value in our model. But the decision rule perversely insists that we must refuse to reject the null when the observation has negative value which happens only when the null hypothesis is false. So, the probability of failing to reject the null hypothesis when in fact the null hypothesis is false (which is what a Type II error is) must be $100\%$ too, right?
Thus, both the Type I error probability and the Type II error probability have value 1 for this (admittedly contrived) example of a decision rule, and so their arithmetic sum must be 2.

Hopefully, the above is enough of a "basic computation of this fact" as the OP desires, or is necessary to resort to the Peano axioms to prove that $1+1=2$?
Example 3: The observation $X \sim U[-2,1]$ whenever $H_0$ is true while  $X \sim U[-1,2]$ whenever $H_a$ is true.  The decision rule is
$$X \begin{array}{c}H_0\\\gtrless\\{H_a}\end{array} 0$$
leading to both the Type I and Type II error probabilities having value $\frac{2}{3}$ and so their sum is $\frac 43$.
Example 4: The observation $X \sim U[-2,1]$ whenever $H_0$ is true while  $X \sim U[-1,2]$ whenever $H_a$ is true.  But now the decision rule is
$$X \begin{array}{c}H_a\\\gtrless\\{H_0}\end{array} 0$$
which is more sensible than the decision rule in Example 3, and it leads to both the Type I and Type II error probabilities having value $\frac{1}{3}$ and so their sum is $\frac 23 < 1$.
A: It is true that with a standard hypothesis test you either reject the null hypothesis or you do not. I.e. type II error + power = 1 under $H_A$ and non-rejection probability + type I error = 1 under $H_0$.
However, the statement the way you phrase it is not true. Type I and type II errors cannot happen under the same scenario within the traditional frequentist hypothesis testing paradigm. I.e. either $H_0$ is true, in which can you can either wrongly reject (type I error) or not reject the null hypothesis, or $H_a$ is true, in which case you can either correctly reject $H_0$ (how often you do this on average is the power) or wrongly not reject (type II error).
