Posted and solved here

Let $a,b$ be those numbers. We need have $ab = (1+2+\cdots + k) - a - b$, or $(a+1)(b+1) = k(k+1)/2 + 1$. Thus we need find the first $k$ such that $k(k+1)/2 + 1$ factorizes such that both factors are between $2$ and $k+1$. Compute the table $\begin{array} {r r r r r r r r r r} k & | & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ k(k+1)/2 + 1 & | & 7 & 11 & 16 & 22 & 29 & 37 & 46 & 56 \\ \end{array}$ to find $k=10$ and $\{a,b\} = \{6,7\}$. EDIT. Well, as I posted the answer I saw the link provided and checked its solution - no wonder going along identical lines. I will let this stay however, if only for saving the trouble to move to the other link. The issue of identifying all such $k$ is probably beyond normal possibilities, since the factors of $k^2+k+2$ cannot be determined in general. However, can we find an infinite family of solutions? The answer is yes - we will take advantage of the particular solution $7\cdot 8 = \dfrac {10^2 + 10 + 2} {2}$, and look for solutions of the form $m(m+1) = \dfrac {k^2 + k + 2} {2}$. This writes as $(2k+1)^2 - 2(2m+1)^2 = -9$, a Pell-like equation with particular solution $(2k+1,2m+1) = (21,15)$. Since the general solution of the fundamental Pell equation $X^2 - 2Y^2 = 1$ is $(x_n,y_n)$, where $x_n + y_n\sqrt{2} = (3 + 2\sqrt{2})^n$, an infinite family is provided by $(2k_n+1,2m_n+1)$, where $(2k_n+1) + (2m_n+1)\sqrt{2} = (21 + 15\sqrt{2})(3 + 2\sqrt{2})^n$.