Problem 261: Pivotal Square Sums

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Let us call a positive integer k a square-pivot, if there is a pair of integers m > 0 and n ≥ k, such that the sum of the (m+1) consecutive squares up to k equals the sum of the m consecutive squares from (n+1) on:

(k-m)2 + ... + k2 = (n+1)2 + ... + (n+m)2.

Some small square-pivots are 4: 32 + 42 = 52 21: 202 + 212 = 292 24: 212 + 222 + 232 + 242 = 252 + 262 + 272 110: 1082 + 1092 + 1102 = 1332 + 1342Find the sum of all distinct square-pivots ≤ 1010.