Saint Peterburg 2016 9.7 wrote: A sequence of $N$ consecutive positive integers is called good if it is possible to choose two of these numbers so that their product is divisible by the sum of the other $N-2$ numbers. For which $N$ do there exist infinitely many good sequences? Answer. $\boxed{\text{All even numbers } N\geq 4.}$ Obviously, $N>2$. Construction for even numbers $N\geq 4$. Let our consecutive numbers be $a+1, a+2, a+3, \ldots, a+N$. We claim that divisibility $$a(N-2)+\frac{N(N+1)}{2}-N-\frac{N}{2}\mid (a+\frac{N}{2})(a+N)$$holds for infinitely many values of $a$. The divisibility simplifies to $$N-2\mid a+N$$and therefore $a=(N-2)k-N$, where $k$ can have infinitely many values. Now we show that all odd numbers $N\geq 3$ do not work. Let our consecutive numbers be $a-\frac{N-1}{2}, a-\frac{N-3}{2}, \ldots, a+0, \ldots, a+\frac{N-3}{2}, a+\frac{N-1}{2}$. For the sake of contradiction assume that the following holds infinitely many times, $$a(N-2)-i-j\mid (a+i)(a+j).$$This actually simplifies to $$a(N-2)-i-j\mid ij(N-2)^2+(i+j)^2(N-1).$$Now, this must hold $(i+j)^2(N-1)=-ij(N-2)^2$, otherwise we would have a finite number of values of $a$. This simplifies to $$i^2(N-1)+i(j(N^2-2N+2))+j^2(N-1)=0\implies i=\frac{-j(N^2-2N+2)\pm\sqrt{j^2(N^2-2N+2)^2-4(N-1)^2j^2}}{2(N-1)}$$Simplifying this even further we get that either $i=-j\cdot (N-1)$ or $i=-j\cdot \frac{1}{N-1}$. This essentially means that $N-1\mid i \text{ or }j$, therefore $i=j=0$, but this also violates the conditions. We are done.