"Reversible cyclic device" is working in Carnot's cycle. The combined system gets an input of heat $\delta Q_R$ that the cycle device turns in part into work - $\delta W_{rev}$ and then gives away residual heat $\delta Q$ towards the "system" which again turns a part of $\delta Q$ into work $\delta W_{sys}$. Now we can write the first law of thermodynamics for the combined system (noting that $\delta W_C=\delta W_{rev}+\delta W_{sys}$) : $$\delta W_C=\delta Q_R-dU_C$$
where $U_C$ is the internal energy of the combined system. From here, using the property of Carnot cycle $$\delta Q_R=\delta Q \frac{T_R}{T}$$ and knowing that there is no net change of internal energy in a cyclic process we obtain $$W_C=T_R \oint \frac{\delta Q}{T}$$
By 2nd law of thermodynamics, we obtain that $W_C$ has to be negative since the system has only one heat source. However, if that is true, then $\delta Q$ has to be negative as well, but is it? It is leaving the cyclic device so it is negative relative to it. However it is entering the "system" so from its perspective it is positive. What am I missing here?