The parity of the sum of the numbers on the board is invariant and hence the final number is odd and also it is positive. I'll assume that the non-negative difference replaces the two numbers $1$ is achievable $[(n+4)^2-(n+3)^2]-[(n+2)^2-(n+1)^2]=4$ and we can make $0$ with $8$ consecutive numbers We make six $4's$ starting from $6^2$ and then make $0$ using the $9\cdot 8$ numbers $29^2,....,101^2$ Now $4^2-(3^2+2^2+1^2)-(5^2-6\cdot 4)=1$