$\sum\limits_{k=1}^{2026} \sin^{-1}\left(\cos\dfrac{k\pi}{4}\right) =$

Let $f(k) = \sin^{-1}\left(\cos\dfrac{k\pi}{4}\right)$.

The function $\cos\dfrac{k\pi}{4}$ is periodic with a period of $8$.

Let us evaluate the sum of $f(k)$ over one complete period, from $k=1$ to $k=8$:

$f(1) = \sin^{-1}\left(\cos\dfrac{\pi}{4}\right) = \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$

$f(2) = \sin^{-1}\left(\cos\dfrac{2\pi}{4}\right) = \sin^{-1}(0) = 0$

$f(3) = \sin^{-1}\left(\cos\dfrac{3\pi}{4}\right) = \sin^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4}$

$f(4) = \sin^{-1}\left(\cos\dfrac{4\pi}{4}\right) = \sin^{-1}(-1) = -\dfrac{\pi}{2}$

$f(5) = \sin^{-1}\left(\cos\dfrac{5\pi}{4}\right) = \sin^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4}$

$f(6) = \sin^{-1}\left(\cos\dfrac{6\pi}{4}\right) = \sin^{-1}(0) = 0$

$f(7) = \sin^{-1}\left(\cos\dfrac{7\pi}{4}\right) = \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$

$f(8) = \sin^{-1}\left(\cos\dfrac{8\pi}{4}\right) = \sin^{-1}(1) = \dfrac{\pi}{2}$

Summing these values, we get:

$\sum\limits_{k=1}^{8} f(k) = \dfrac{\pi}{4} + 0 - \dfrac{\pi}{4} - \dfrac{\pi}{2} - \dfrac{\pi}{4} + 0 + \dfrac{\pi}{4} + \dfrac{\pi}{2} = 0$

Since the sum over any $8$ consecutive terms is $0$, we divide the total number of terms by $8$:

$2026 = 8 \times 253 + 2$

This means there are $253$ complete periods whose sum is $0$, leaving only the first $2$ terms of the next period.

$\sum\limits_{k=1}^{2026} f(k) = 253 \times 0 + f(1) + f(2)$

$\sum\limits_{k=1}^{2026} f(k) = \dfrac{\pi}{4} + 0 = \dfrac{\pi}{4}$

Answer: $\dfrac{\pi}{4}$