jmc

We compute directly using the given recursive definition: $$\begin{array}{rl}f(94)& =94^2-f(93)\\ & =94^2-93^2+f(92)\\ & =94^2-93^2+92^2-f(91)\\ & =\cdots \\ & =94^2-93^2+92^2-91^2+\cdots +20^2-f(19)\\ & =(94^2-93^2+92^2-91^2+\cdots +20^2)-94.\end{array}$$To compute this sum, we write $$\begin{array}{rl}94^2-93^2+92^2-91^2+\cdots +20^2& =(94^2-93^2)+(92^2-91^2)+\cdots +(22^2-21^2)+20^2\\ & =(94+93)+(92+91)+\cdots +(22+21)+20^2\\ & =\frac{1}{2}(94+21)(94-21+1)+400\\ & =4255+400\\ & =4655.\end{array}$$Therefore, $$f(94)=4655-94=\boxed{4561}.$$