jmc

Let $S=g(1)+g(2)+\cdots +g(20).$ Then by definition of a tenuous function, $$\begin{array}{rl}S& =[g(20)+g(1)]+[g(19)+g(2)]+[g(18)+g(3)]+\cdots +[g(11)+g(10)]\\ & \ge (20^2+1)+(19^2+1)+(18^2+1)+\cdots +(11^2+1)\\ & =2495\end{array}$$Let's assume that $S=2495,$ and try to find a function $g(x)$ that works. Then we must have $$\begin{array}{rl}g(20)+g(1)& =20^2+1,\\ g(19)+g(2)& =19^2+1,\\ g(18)+g(3)& =18^2+1,\\ & \dots ,\\ g(11)+g(10)& =11^2+1.\end{array}$$If $g(1)<g(2),$ then $$g(19)+g(1)<g(19)+g(2)=19^2+1,$$contradicting the fact that $g$ is tenuous.

And if $g(1)>g(2),$ then $$g(20)+g(2)<g(20)+g(1)=20^2+1,$$again contradicting the fact that $g$ is tenuous. Therefore, we must have $g(1)=g(2).$

In the same way, we can prove that $g(1)=g(3),$ $g(1)=g(4),$ and so on, up to $g(1)=g(10).$ Hence, $$g(1)=g(2)=\cdots =g(10).$$Let $a=g(1)=g(2)=\cdots =g(10).$ Then $g(n)=n^2+1-a$ for all $n\ge 11.$ Since $g(11)+g(11)\ge 122,$ $g(11)\ge 61.$ But $g(11)=121+1-a=122-a\le 61,$ so $a\le 61.$ The smallest possible value of $g(14)$ is then $14^2+1-61=\boxed{136}.$