jmc

Setting $x=\pi ,$ we get $$(-1{)}^n\ge \frac{1}{n},$$so $n$ must be even. Let $n=2m.$

Setting $x=\frac{\pi }{4},$ we get $${\left(\frac{1}{\sqrt{2}}\right)}^{2m}+{\left(\frac{1}{\sqrt{2}}\right)}^{2m}\ge \frac{1}{2m}.$$This simplifies to $$\frac{1}{2^{m-1}}\ge \frac{1}{2m},$$so $2^{m-2}\le m.$ We see that $m=4$ is a solution, and the function $2^{m-2}$ grows faster than $m,$ so $m=4$ is the largest possible value of $m.$

We must then prove that $${\sin }^{8}x+{\cos }^{8}x\ge \frac{1}{8}$$for all real numbers $x.$

By QM-AM, $$\sqrt{\frac{{\sin }^{8}x+{\cos }^{8}x}{2}}\ge \frac{{\sin }^{4}x+{\cos }^{4}x}{2},$$so $${\sin }^{8}x+{\cos }^{8}x\ge \frac{({\sin }^{4}x+{\cos }^{4}x{)}^2}{2}.$$Again by QM-AM, $$\sqrt{\frac{{\sin }^{4}x+{\cos }^{4}x}{2}}\ge \frac{{\sin }^{2}x+{\cos }^{2}x}{2}=\frac{1}{2},$$so $${\sin }^{4}x+{\cos }^{4}x\ge \frac{1}{2}.$$Therefore, $${\sin }^{8}x+{\cos }^{8}x\ge \frac{(1/2{)}^2}{2}=\frac{1}{8}.$$We conclude that the largest such positive integer $n$ is $\boxed{8}.$