1999 AIME Problem 5

If adding $2$ causes no carrying, $T(x) = 2.$ Carrying happens only when $x$ ends in $8$ or $9.$

Track how the digit sum changes when $x$ ends in a run of nines; the carrying values form an arithmetic progression with common difference $9$

If the last digit of $x$ is at most $7,$ adding $2$ changes no other digit, so $T(x) = 2.$ Otherwise there is carrying. If $x$ ends in the digit $8$ preceded by exactly $m \ge 0$ nines, then $x + 2$ replaces $\ldots a\underbrace{9 \cdots 9}_{m}8$ by $\ldots (a{+}1)\underbrace{0 \cdots 0}_{m}0,$ so $S(x + 2) - S(x) = 1 - 9m - 8$ and $T(x) = 9m + 7.$ If $x$ ends in exactly $m \ge 1$ nines, then $x + 2$ replaces $\ldots a\underbrace{9 \cdots 9}_{m}$ by $\ldots (a{+}1)\underbrace{0 \cdots 0}_{m - 1}1,$ so $S(x + 2) - S(x) = 2 - 9m$ and $T(x) = 9m - 2.$

Both carrying families give exactly the values $7, 16, 25, \ldots,$ that is, $9j + 7$ for $j \ge 0,$ and every such value occurs. So the possible values of $T$ are $2$ together with all $9j + 7.$ Requiring $9j + 7 \le 1999$ gives $j \le 221,$ which is $222$ values, and $T = 2$ adds one more, for a total of $223.$