, 2019

Let $18 = n$, so that the initial message has length $4n$, with exactly $n$ symbols of each kind.
We first establish an invariant. Assign
$E = 3$, $X = -3$, $I = 2$, $T = -2$,
and let $S$ denote the sum of the values of all symbols appearing in the message. Initially, $S = 0$, and the sum stays invariant under all the legal transformations.

Our next observation is that we may always insert or delete the combination TETET, for
$$\emptyset \mapsto \{TI\} \mapsto \{TEXI\} \mapsto \{TETIXI\} \mapsto \{TETEXIXI\} \mapsto \{TETETIXIXI\} \mapsto \{TETET\},$$
and this works also in reverse. Required are six operations.

To crack the system, first insert XE behind every I, and XE in front of every T:
$I \mapsto IXE$, $\quad$ $T \mapsto XET$.
This requires at most $p_1 = 4n$ operations, after which the message has length at most $12n$. It may now be considered a sequence of the four possible strings
$E$, $X$, $IX$, $ET$.

Second, expand any single $X$ (not preceded by an $I$) into $IXIXIX$, and any single $E$ (not succeeded by a $T$) into $ETETET$:
$X \mapsto IXIXIX$ (one operation), $E \mapsto ETETET$ (six operations).
This requires at most $p_2 = 12n \cdot 6 = 72n$ operations, and the message now has length at most $72n$.
It is at present reduced to some binary combination of the two strings
$IX$, $ET$.

Third, effectuate all possible reductions
$$\{ETIX\} \mapsto \{EX\} \mapsto \emptyset \quad \text{and} \quad \{IXET\} \mapsto \{IT\} \mapsto \emptyset.$$
There can be at most $18n$ such reductions, totalling $p_3 = 18n \cdot 2 = 36n$ operations. The hacker will be left with a message of the types
$ETET \dots ET$ or $IXIX \dots IX$
or an empty screen. But since the invariant $S = 0$, the screen must now, in fact, be empty.

Fourth, insert EXIT, using $p_4 = 2$ more operations. The number of operations was at most
$$p_1 + p_2 + p_3 + p_4 = 4n + 72n + 36n + 2 = 112n + 2 = 112 \cdot 18 + 2 = 2018 < 2019.$$