jmc

Let the other 2010 numbers be $y_1,$ $y_2,$ $\dots ,$ $y_{2010}.$ Then $y_1+y_2+\cdots +y_{2010}=2012-x$ and $\frac{1}{y_1}+\frac{1}{y_2}+\cdots +\frac{1}{y_{2010}}=2012-\frac{1}{x}.$ By Cauchy-Schwarz, $$\left(\sum _{i=1}^{2010}{y}_i\right)\left(\sum _{i=1}^{2010}\frac{1}{y_i}\right)=(2012-x)\left(2012-\frac{1}{x}\right)\ge 2010^2.$$Then $2012^2-2012\left(x+\frac{1}{x}\right)+1\ge 2010^2,$ which leads to $$x+\frac{1}{x}\le \frac{8045}{2012}.$$The equation $x+\frac{1}{x}=\frac{8045}{2012}$ reduces to $x^2-\frac{8045}{2012}x+1=0,$ which has real roots. We can then set $y_i=\frac{2012-x}{2010}$ in order to achieve equality. Thus, the maximum value is $\boxed{\frac{8045}{2012}}.$