I try to introduce a roundabout production as (Basu 1995) in the new keynesian model. The model run when there is only labor in production function. When I add the intermediate input in the production function I get this error message:
Error using print_info (line 78)

The steady state is complex
Error in steady (line 92)
print_info(info,options_.noprint, options_);
Error in modelLin_3 (line 338)
steady;
Error in dynare (line 180)
evalin(‘base’,fname) ;

Jt is the lagrange multiplier for employment law of motion and can be interpreted as the value of marginal worker to a firm. If I set phi_c=0.00001, the production function include only the labor nt and Jt=phitAt-(1-gamma+gammaRt)wt+(1-rho)betaEt(lamt+1/lamt)Jt+1. When I introduce intermediate input in the production function,
Jt=phitAt(1-phi_c)*sig^(phi_c)n^(-phi_c)-(1-gamma+gammaRt)*wt+(1-rho)betaEt(lamt+1/lamt)*Jt+1. without input Js is positive and results are in line with author. When I have input intermediate Js turn into negative. I recalculated steady state (by hand) and I have the same problem. What does that mean? The model I propose I not resolvable or I have to change calibration for example?

That is hard to tell. I am not familiar with your model, but in many models the Lagrange multiplier has an economic interpretation. For example, the multiplier attached to the budget constraint measures marginal utility of wealth. If it became negative, the household would be better off if it had a tiny bit less wealth, which economically does not make sense.
If a similar restriction holds in your model that the Lagrange multipliers should be positive, you either have a mistake in your derivations or the calibration is problematic.

I rewrite the model with the new derivation and still have a problem, the error message now is
Error using print_info (line 51)
The Jacobian matrix evaluated at the steady state contains elements that are not
real or are infinite
Error in check (line 76)
print_info(info, 0, options);
Error in model_lininput4 (line 335)
oo_.dr.eigval = check(M_,options_,oo_);
Error in dynare (line 180)
evalin(‘base’,fname) ;
Error in excutelininput4 (line 72)

Check your equations again. Given your starting values, equation (13) still has a residual. The steady state then found features a negative sig that results in taking the root of a negative number in in the derivative of equation (13) in