Making it a Perfect Square

Let $x$ be the smallest element of such a set $S$ ; then we have $\sum_{s \in S} s = \sum_{i = 0}^{125} x + 2 * i = 126x + 2 \sum_{i = 0}^{125} i = 126x + 126 \cdot 125 = 126(x + 125).$ For this sum to be a perfect square, each of its prime factors must appear an even number of times in the prime factorization. So far we see the sum is divisible by $126 = 2 \cdot 3^2 \cdot 7$ , so the smallest perfect square the sum could equal is $2^2 \cdot 3^2 \cdot 7^2$ . To achieve this, we need the $x + 125$ term to provide additional powers of 2 and 7 so that their powers are even. In particular, since we are looking for the smallest such $x$ , we choose $x + 125 = 2 \cdot 7 = 14$ . From here we derive $x = -111$ , and so the largest element of $S$ is $x + 2 \cdot 125 = \boxed{139}$ .