9 + 16 = 25 is one of them

Using the identity $\displaystyle\sum_{m=1}^{N} m^{2} = \dfrac{N(N + 1)(2N + 1)}{6}$ the given equation can be written as

$\displaystyle\sum_{m=1}^{n+k} m^{2} - \sum_{m=1}^{n-1} m^{2} = \sum_{m=1}^{n + 2k} m^{2} - \sum_{m=1}^{n + k} m^{2} \Longrightarrow 2\sum_{m=1}^{n+k}m^{2} - \sum_{m=1}^{n-1} m^{2} = \sum_{m=1}^{n + 2k} m^{2}$

$\Longrightarrow \dfrac{2(n + k)(n + k + 1)(2n + 2k + 1)}{6} - \dfrac{(n - 1)(n)(2n - 1)}{6} = \dfrac{(n + 2k)(n + 2k + 1)(2n + 4k + 1)}{6}.$

Expanding and simplifying yields the quadratic $n^{2} - 2k^{2}n - k^{2}(1 + 2k) = 0$ , for which

$n = \dfrac{2k^{2} \pm \sqrt{4k^{4} + 4k^{2}(1 + 2k)}}{2} = k^{2} \pm k\sqrt{k^{2} + (1 + 2k)} = k^{2} \pm k\sqrt{(k + 1)^{2}} = k^{2} \pm k(k + 1).$

Now as we are looking for positive $n$ we need only consider the positive root

$n = k^{2} + k(k + 1) = k(2k + 1).$

Finally, as $31(2*31 + 1) = 1953 \lt 2016$ and $32(2*32 + 1) = 2080 \gt 2016$ we will find suitable values for $n$ for each integer $k$ such that $1 \le k \le 31$ , and so the desired answer is $\boxed{31}.$

To simplify the computations a bit, we can use symmetry and let $n+k=m$ ; the equation now takes the form $(m-k)^2+...+(m-1)^2+m^2=(m+1)^2+...+(m+k)^2$ or $(k+1)m^2-2m\sum_{j=1}^{k}j+\sum_{j=1}^{k}j^2=km^2+2m\sum_{j=1}^{m}j+\sum_{j=1}^{k}j^2$ or $m=4\sum_{j=1}^{k}j=2k(k+1)$ so that $n=m-k=k(2k+1)$ . The equation $k(2k+1)=2016$ has the positive solution $k=31.5$ so that we can let $1\leq k \leq 31$ , giving $\boxed{31}$ solutions.