31st Hellenic Mathematical Olympiad

The given relation is equivalent to: $S_1=\frac{39}{103}(S_1+S_2)$. Since $$S_1+S_2=1+2+\cdots +n^2=\frac{n^2(n^2+1)}{2}$$ and $S_1$ is a natural number we conclude that $103\mid \lfloor \frac{n^2(n^2+1)}{2}\rfloor$. Since $103$ is a prime number of the form $4\kappa +3$, it follows that $103$ does not divide $n^2+1$. Hence the prime number $103$ must divide $n^2$ and finally $103|n$. Moreover $n$ is even, and so: $n=206k,k\in {\mathbb{N}}^{\ast }$.

Now we are going to prove that for every number $n=206k,k\in \mathbb{Z}$, it is possible to obtain the wanted arrangement. The least (maximal) possible value of $S_1$ is: $$1+2+\cdots +\frac{n^2}{2}=\frac{\frac{n^2}{2}\left(\frac{n^2}{2}+1\right)}{2}=A,\left(\frac{n^2}{2}+1\right)+\left(\frac{n^2}{2}+1\right)+\cdots +n^2=B,$$ respectively. Easily we can check the inequality, $A<S_1=\frac{39}{103}(S_1+S_2)<B$. Our result arrives by proving that $S_1$ can obtain every possible value between $A,B$. In fact, number $A+1$ can be obtained by putting in the white squares the numbers $1,2,\dots ,\frac{n^2}{2}-1,\frac{n^2}{2}+1$. The number $A+2$ can be obtained by putting in the white squares the numbers $1,2,\dots ,\frac{n^2}{2}-1,\frac{n^2}{2}+2$, and so on. When we arrive to the step we need to put the numbers $1,2,\dots ,\frac{n^2}{2}-1,\frac{n^2}{2}$, in order to obtain the next number for their sum, then we select in the white squares the numbers $1,2,\dots ,\frac{n^2}{2}-2,\frac{n^2}{2}$, $n^2$, and for the next number we select $1,2,\dots ,\frac{n^2}{2}-2,\frac{n^2}{2}+1,n^2$, and so on. Hence, using the above procedure to increase at every step the sum by $1$, starting from $A$, we can construct all integers till $B$.