Add some integers to a multiset so that $\sum\arctan\frac1x=\frac\pi2$ 
$S$ is a finite multiset of positive integers for which $\displaystyle\sum_{x\in S}\arctan\frac1x<\frac\pi2$. Show there $\exists S’,S\subset S’\displaystyle\sum_{x\in S’}\arctan\frac1x=\frac\pi2$.

Let $S=\{a_1,a_2,\ldots,a_n\}$. We add $b_1,b_2,\ldots,b_t$ to $S$ to form $S’$. For each item $s$ in $S’$ we construct a corresponding complex number $s+i$ because $\arg(s+i)=\arctan\frac1s$.
Using complex number property can get $\displaystyle\prod_{k=1}^n(a_k+i)\prod_{j=1}^t(b_j+i)\in i\mathbb R$. Perhaps this is now a polynomial problem: the sum of coefficients of terms with exponent $\equiv2\pmod4$ and $\equiv0\pmod4$ are the same.

Using the formula $\tan(x+y)=(\tan x+\tan y)/(1-\tan x\tan y)$, it is not difficult to check that the sum of $\arctan1/x$ is $\arctan a/b$ with positive integers $a,b$. Adding many $\arctan1/M$ with large $M$, we can reach $a>b$. Now, we have
$$
 \arctan\frac ab+\arctan\frac 1N=\arctan\frac{\frac ab+\frac 1N}{1-\frac{a}{Nb}}=\arctan\frac{Na+b}{Nb-a}.
$$
Choosing the first $N$ such that $Nb-a>0$, we obtain
$$
 \arctan\frac ab+\arctan\frac 1N=\arctan\frac{a_1}{b_1},\ \ \ b_1<b.
$$
Repeating this procedure we arrive at the case $b_k=1$. Thus, the sum of all $\arctan$ in the new set is $\arctan A$ with some positive integer $A$. Finally,
$$
 \arctan A+\arctan \frac1A=\frac\pi2.
$$
