The answer is $4$. Suppose that the five disjoint pairs have common sum $s$ and the $10$ distinct numbers are $x, y, z, t, w, s-x, s-y, s-z, s-t, s-w$. To find an example with $4$ equal products, it's sufficient to find a solution of the system $xy=zt=(s-x)(s-z)=w(s-t)$. Fix common product $12$, achievable by $x=2, y=6, z=4, t=3$. Then $(s-2)(s-4)=s^2-6s+8=12 \iff s^2-6s-4=0$, so $s=3+\sqrt{13}$ and $w=\frac{12}{\sqrt{13}}$ indeed work and the $10$ numbers are $$2, 6, 3, 4, \sqrt{13}+1, \sqrt{13}-1, \sqrt{13}, \sqrt{13}-3, \frac{12}{\sqrt{13}}, 3+\frac{1}{\sqrt{13}}.$$ Now suppose that we can split the $10$ numbers into $5$ pairs with equal products and let again the common sum from the statement be $s$. At most one pair of those five has sum $s$, because otherwise $a(s-a)=b(s-b)$ implies that $s=a+b$ and the pairs $(a, s-a)$ and $(b, s-b)$ actually coincide. Take one of the pairs $(x, y)$ such that $x+y \neq s$. The numbers $s-x, s-y$ should be among the ten numbers and they can’t be in the same pair, as $(s-x)(s-y)=xy$ implies $x+y=s$. Thus, they are in distinct pairs, and suppose they are $(s-x, z)$ and $(s-y, t)$. We have two cases: Case 1. $z+t=s$. Then the first three pairs are $(x, y), (s-x, z), (s-y, s-z)$, so the rest $4$ numbers are of the form $a, b, s-a, s-b$. However, it's easy to see by case inspection that those $4$ numbers can't be split into two pairs with equal products, as they are pairwise distinct. Case 2. $z+t \neq s$. Now, we claim that the numbers $s-z, s-t$ do not appear among $x, y, s-x, s-y, z, t$. It's obvious that none of them can be equal to some of the last $4$ numbers. Since $s-x \neq z$ and $s-y \neq t$, the only possibility is $s-z=y$ or $s-t=x$. But if only one of those holds, $z$ or $t$ will participate in two pairs, so they both hold and we obtain $(s-x)(s-y)=zt=xy$, or that $x+y=s$, contradiction. Hence, the numbers $s-z, s-t$ are "new" and are among the $10$ numbers, so we have two possibilities: either they are in the same pair $(s-z, s-t)$ and the other pair is $(w, s-w)$ for some $w>0$, or they are in distinct pairs, which are $(s-z, w)$ and $(s-t, s-w)$. Case 2.1. We have $xy=(s-x)z=(s-y)t=(s-z)(s-t)=w(s-w)$. Then $z=\frac{xy} {s-x}$ and $t=\frac{xy} {s-y}$, so $(s-z)(s-t)=xy$ rewrites as $(s-x-y)(s^2-2xy)=0$. By the choice of $x, y$, we obtain $s^2=2xy$. But this implies that $w^2-sw+xy=0$ has discriminant $D=s^2-4xy=-2xy < 0$, contradiction, as this means there is no solution for $w$. Case 2.2. We have $xy=(s-x)z=(s-y)t=(s-z)w=(s-t)(s-w)$. Similarly, $z=\frac{xy} {s-x}$ and $t=\frac{xy} {s-y}$, so $s=\frac{xy} {s-z}+\frac{xy}{s-t}$ rewrites as $(s-x-y)(s^4-3s^2xy+x^2y^2)=0$. This implies that $s^2=\frac{3\pm\sqrt{5}}{2}xy$. Since $0<x, y<s$, we have $xy<s^2$ and combined with $\frac{3-\sqrt{5}}{2}<1$ makes the first possibility impossible. Hence, $s^2=\frac{3+\sqrt{5}}{2}xy$. We have $z=\frac{xy} {s-x}<s$, which means that $s^2-sx-xy=\frac{1-\sqrt{5}}{2}xy-sx>0 \iff 0>\frac{1-\sqrt{5}}{2}y>s$, contradiction. As all cases are exhausted, we are done with the proof of the bound. $\textbf{Remark.}$ It was not a very pleasant experience to do this problem during the test, though the case bash approach for the bound kind of helped me to find the construction, which appears to be the harder part according to most of the people (except me). Also, I thank @SevroMyName for finishing Case 2.2.