jmc

Let the first term be $a$, and let the common difference be $d$. Then the four positive integers are $a$, $a+d$, $a+2d$, and $a+3d$. The sum of these four positive integers is $4a+6d=46$, so $2a+3d=23$. Solving for $d$, we find $d=(23-2a)/3$.

The third term is $$a+2d=a+2\cdot \frac{23-2a}{3}=\frac{46-a}{3}.$$ Thus, to maximize this expression, we should minimize $a$. Since $a$ is a positive integer, the smallest possible value of $a$ is 1. Furthermore, when $a=1$, $d=(23-2)/3=7$, which gives us the arithmetic sequence 1, 8, 15, 22. Therefore, the greatest possible third term is $\boxed{15}$.