I think the answer is no. The "t" in a t-test refersT tests, basically, work as follows. You manipulate your input data to obtain something that (under the null hypothesis) is normally distributed with mean zero. For example, if the null hypothesis is that the data is normally distributed with mean 28, then the manipulation is simply subtracting 28 from the data. For more complex cases (e.g. if there are two populations with the same, but unknown, mean), the manipulation is a bit more involved, but still some manipulation is needed that gives you a normal distribution with zero mean. Once you obtain that, you use the fact that $$\frac{N - m}{s}$$$\frac{N}{s}$ (where $s$ is sample standard deviation) is distributed according to Student t (that's where the t distribution. So ain t-test always assumes that there is some fixed mean $m$; it may be unknown, but it should be fixedcomes from) to determine confidence.
In your case, the null hypothesis encompasses many values for the mean of one population. So I don't think a standard t-test would applysee an obvious way to manipulate it into something that has mean zero. For example, if you have paired data and subtract the two populations, you get $X_i - Y_i \sim N(m, \sigma)$ where $m \geq \delta$. You can further subtract $\delta$ to get mean $m \geq 0$. But that still encompasses many possible values of the mean. I don't see how to manipulate it further to get something that has just one possible value of the mean.