A finite division ring is a field and the multiplicative group of a finite field is cyclic. Let $F$ be a finite field of order $q$ and let $g$ be a generator of $F^{\times},$ the multiplicative group of $F.$ So $0=g^{q-1}-1=(g-1)\sum_{k=0}^{q-2}g^k$ and thus, since $g \ne 1,$ $$\sum_{k=0}^{q-2}g^k=0. \ \ \ \ \ \ \ \ \ \ \ \ (*)$$Now consider two cases. Case 1. $q$ is even. In this case we choose $B=\{1\}$ and $A:=F^{\times}\setminus B.$ Then, by $(*),$ $$\sum_{x \in A}x=-1=1=\prod_{y \in B}y.$$ Case 2. $q$ is odd. In this case we choose $B=\{1,-1,g^i\},$ where $g^i \ne \pm 1,$ and $A:=F^{\times}\setminus B,$ which is non-empty because $q \ge 5.$ Now, by $(*),$ $$\sum_{x \in A}x=-\sum_{y \in B}y=-g^i=\prod_{y \in B}y.$$

Why, in case 1, $ q $ is even implies $ \text{char} F=2 $ ?

CatalinBordea wrote: Why, in case 1, $ q $ is even implies $ \text{char} F=2 $ ? characteristic of finite field is a prime number?

Yes, yes. Back then didn't saw that $ q $ is the order. By the way the solution of ysharifi is neater than the standard one.

$\sum_{y \in B}y + \prod_{y \in B}y =0$ this alternative solution allows us to construct B with 2 elements instead of 3. (q odd) $B=\{b_1,b_2\}$ $b_1b_2 +b_1+b_2=0$ $(b_1+1)(b_2+1)=1$ let $(b_1,b_2)=(g-1,g^{q-2}-1)$ this is one possible B but there are $\frac{q-3}{2}$ (q odd) B's with 2 elements. CatalinBordea wrote: By the way the solution of ysharifi is neater than the standard one. so what is the std one? a more challenging problem is how many ways are there to partition the finite division ring into A and B st the eqn is satisfied.

Observe that $ \prod_{x\in K\setminus\{ 0 \}} x =\prod_{\substack{y\in K\setminus\{ 0 \} \\ \text{ord} (y)\le 2 }} y=-1. $ This implies that for any partition $ A\cup B=K\setminus\{ 0 \}, $ we have $ \sum_{x\in A} x=\prod_{y\in B} y $ iff: $$ \left( \sum_{x\in A} x \right)\cdot\left(\prod_{x\in A} x\right) =-1\quad\text{(*)} $$Observe, again, that the set of roots in $ K $ of the polynomial $ f:=X^{|K|-1}\in K[X] $ is the set of elements of $ K\setminus\{ 0 \} , $ so, the set of roots in $ K $ of any divisor of $ f $ has a number of elements equal to the degree of this divisor. $ \text{(**)} $ Case 1: If $ \text{char} K=2, $ then $ A=\{ 1\} $ satisfy $ \text{(*)} . $ Case 2: If $ \text{char} K>2 $ and $ |K|\equiv 1\pmod 4, $ then, $ X^2+1| \left( X^2+1 \right)\left( X^2-1 \right) = X^4-1| f $ so, there exists an $ x\in K\setminus\{ 0 \} $ such that $ x^2=-1. $ Now, $ A=\{ x \} $ satisfy $ \text{(*)} . $ Case 3: If $ \text{char} K>2 $ and $ |K|\not\equiv 1\pmod 4, $ then also by the hypothesis, $ |K|-1 $ is even and greater than $ 6, $ which ensures the existence of an odd divisor $ d>1 $ of $ |K|-1. $ Noting, that $ 1+X+X^2+\cdots +X^{d-1} | f, $ according to $ \text{(**)} , $ we obtain a set of $ d-1 $ distinct elements of $ K\setminus\{ 0 \} $ and, due to Viète's relations and the fact that $ d-1 $ is even, the sum of these roots is equal to $ -1 $ while their product is $ 1 $ and we chose $ A $ to be this set in order to satisfy $ \text{(*)} $ and finish the problem.

i dont think A is a ring. maybe you meant char(K)?

Yes, I meant that.

Finite division rings are finite fields. Let the field be $K$ and let $|K|=:q$. Since $K$ is a domain, $x^2-1$ has at most $2$ roots. If $K$ has characteristic $2$ then $x^2-1$ has one root $x=1$. Thus every $a\in K^\times\setminus\{1\}$ has a unique multiplicative inverse not equal to itself, so taking $A:=\{1\}$ and $B:=K^\times\setminus\{1\}$ works. So assume $K$ has characteristic $\geq 3$. Then $x^2-1$ has two roots $1$ and $-1$. Hence every $a\in K^\times\setminus\{1,-1\}$ has a unique multiplicative inverse not equal to itself, so \[ \prod_{a\in K^\times}a=-1. \]If $q-1$ is a power of $2$, then $4\mid q-1$ since $q\geq 4$. Since $K^\times$ is cyclic, there exists $g\in K^\times$ of order $4$. In particular, $g^2=-1$. Thus, taking $A:=\{g\}$ and $B:=K^\times\setminus\{g\}$ works: \[ g=-g^{-1}=\prod_{b\in B}b. \]So assume $q-1$ has an odd prime factor $p$. Then $K^\times$ has an element $g$ of order $p$. Note that since $g\neq 1$, \[ (g-1)(g^{p-1}+\cdots+g+1)=g^p-1=0 \]so $g^{p-1}+\cdots+g+1=0$. Thus, taking \[ A:=\left\{g^i\ \Big|\ i\in\mathbb{Z}\setminus\{0\},-\frac{p-1}{2}\leq i\leq\frac{p-1}{2}\right\} \](the $g^i$ are distinct since $\mathrm{ord}(g)=p$) and $B:=K^\times\setminus A$ works: \[ \sum_{a\in A}a=g^{-\frac{p-1}{2}}\left(g^{p-1}+\cdots+g+1-g^{\frac{p-1}{2}}\right)=g^{-\frac{p-1}{2}}\left(-g^{\frac{p-1}{2}}\right)=-1=-\left(\prod_{a\in A}a\right)^{-1}=\prod_{b\in B}b. \]We are done. $\square$