Consecutive Square-Infected Positive Integers

Actually,

For every positive integer $k$ , there are $k$ consecutive positive integers, none of which is square-free.

Let's demonstrate a constructive proof.

Proof: Solve the following system of $k$ linear congruences using the Chinese Remainder Theorem : $x \equiv -i (\mod p_i^2) \text{, for } 1\leq i \leq k,$ where each $p_i$ is distinct prime. Assume the solution is: $x\equiv X (\mod M)$ , where $M=\prod_1^k p_i^2$ , and $X$ is a non-negative integer.

Then, $(X+1),(X+2),\cdots{ }\cdots{ } \cdots{ }$ and $(X+k)$ are $k$ consecutive positive integers, none of which is square-free, because for every $i$ (with $1\leq i \leq k$ ), $p_i^2 \mid (X+i)$ . $\blacksquare$

So the sum of all possible values of $k = 1+2+3+\cdots{ }\cdots{ } \cdots{ }+2017=\frac{2017\times 2018}{2}=\boxed{2035153}$ .