On account of the fact that there is interesting mathematics at play here, I will answer this question. Please do not construe this as medical advice.
Without knowing what level of presence is considered "out of one's system", this question cannot be given a full answer. However, I can give an expression for the amount of substance that will remain after time $t$. All units will be in hours and milligrams. Since the half-life is $37$, the mass of substance $m(t)$ as a function of time satisfies the differential equation $$\frac{d}{dt}m(t)=\frac{\ln 2}{37}m(t)$$
and in addition we have discontinuities introduced each day for the first month by the fact that $m(t)$ increases by $2$ at 24-hour intervals for the first month. At time $0$, the first dose is taken, so $m(0)=2$. Observing that the solution to the differential equation away from the discontinuities is $$m(t_0)e^{-\lambda(t-t_0)}$$
where $\lambda = \frac{\ln 2}{37}$ we get that $m(24) = 2 + 2e^{-24\lambda}$. Repeating in this manner,
$$m(48)=2+m(2)e^{-24\lambda} = 2 + (2 + 2e^{-24\lambda}) e^{-24\lambda} = 2+2e^{-24\lambda}+2e^{-48\lambda}$$
and more generally
$$m(24n) = 2\sum_{k=0}^n e^{-24k\lambda}$$
for $n\leq 30$, assuming that the month in question has 31 days. After $24\times 30=720$ hours, no more doses will be taken and so we will be away from the discontinuities, thus for $t>720$ we have
$$m(t)=m(720)e^{-\lambda(t-720)}=2\left(\sum_{k=0}^{30} e^{-24k\lambda}\right)e^{-\lambda(t-720)}$$
which can be graphed by most mathematics programs, including wolfram alpha (keep in mind that $\lambda$ is a constant with approximate value $.0187$). Once this graph dips below the relevant threshold value, the substance will be "out of one's system".