Let's try the alternative approach of formalizing, and actually calculating the answer, and thereby discover that there are really two different answers, depending how you interpret the problem.
As in many cases, the most important point is how to formalize the problem. The OP and several earlier answers use variables for propositions like 'Steve has the highest grade', etc. Instead, it seems to me a lot simpler to use variables for the grades themselves: $\;s,h,j\;$ stand for Steve's, Helen's, and Jack's grades respectively.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
$
Now we can easily formalize 'Steve has the highest grade' as $\;h \le s \land j \le s\;$, 'Jack has the lowest grade' as $\;j \le s \land j \le h\;$, etc.
Note an important point here: I chose to use $\;\le\;$ instead of $\;\lt\;$, since the problem does not clearly say or imply that all the three grades are different: perhaps one of them has the lowest grade, and both others the highest?
With this formalization, we are given two things:
\begin{align}
\tag{1}
\lnot(h \le s \land j \le s) \;\then\; h \le j \land s \le j
\\
\tag{2}
\lnot(j \le s \land j \le h) \;\then\; s \le h \land j \le h
\end{align}
and we are asked to determine the relative order of $\;s,h,j\;$.
As an earlier answer already suggests, it is often easier to write 'then' ($\;\then\;$) in terms of 'or' ($\;\lor\;$), using $\;\phi \then \psi \;\equiv\; \lnot \phi \lor \psi\;$. Doing that results in the equivalent and more symmetrical
\begin{align}
\tag{1a}
(h \le s \land j \le s) \;\lor\; (h \le j \land s \le j)
\\
\tag{2a}
(j \le s \land j \le h) \;\lor\; (s \le h \land j \le h)
\end{align}
Looking at these, it seems that $\Ref{2a}$ is easiest to simplify, because both sides contain $\;j \le h\;$:
$$\calc
\tag{2a}
(j \le s \land j \le h) \;\lor\; (s \le h \land j \le h)
\op\equiv\hint{logic: $\;\land\;$ distributes over $\;\lor\;$}
(j \le s \lor s \le h) \;\land\; j \le h
\op\equiv\hints{left hand part follows from $\;j \le h\;$, since $\;\le\;$ is a}\hint{total order}
j \le h
\tag{2b}
\endcalc$$
Now we can combine all and simplify further:
$$\calc
\Ref{1} \;\land\; \Ref{2}
\op\equiv\hint{use $\Ref{1a}$ and $\Ref{2b}$}
((h \le s \land j \le s) \;\lor\; (h \le j \land s \le j)) \;\land\; j \le h
\op\equiv\hint{logic: $\;\land\;$ distributes over $\;\lor\;$}
(h \le s \land j \le s \land j \le h) \;\lor\; (h \le j \land s \le j \land j \le h)
\op\equiv\hint{simplify}
j \le h \le s \;\lor\; s \le j = h
\endcalc$$
And we have found our ordering(s).
Interpreting this result, we see that there are two possibilities: either Jack has the lowest grade and Steve has the highest; or Jack and Helen both have the highest grade.
(And if there are no ties, then that of course prevents the second possibility.)
