If $$x_1, x_2, \ldots, x_{49}$$ are non-zero integers such that $$\sum_{i=1}^{49} x_i = 0$$, then what is the minimum possible value of $$\sum_{i=1}^{49} x_i^2$$?

Every non-zero integer has square at least 1, but the 49 numbers cannot all be $$1$$ or $$-1$$, since a sum of an odd number of odd integers is odd and so cannot be 0. At least one number therefore has absolute value at least 2, giving $$\sum_{i=1}^{49} x_i^2 \ge 48 \times 1 + 2^2 = 52$$. This bound is attained by 25 copies of 1, 23 copies of $$-1$$ and one $$-2$$, since $$25 - 23 - 2 = 0$$ and $$25 + 23 + 4 = 52$$.