Let $P(i)$ denote the probability he'll get to $\$1,000,002$ if he starts from $\$i$. We know that $$P(0)=0\;\;\;P(1,000,002)=1$$ You are asked to find $P(1,000,000)$.
Imagine he has $\$i$ and places a bet. He either gets to state $(i+1)$ or to state $(i-1)$ and we have: $$P(i) = \frac 13 P(i+1)+\frac 23 P(i-1)$$ It is easier to start near the "ruin state". We see, for example, that $$P(1) = \frac 13 P(2)+\frac 23 P(0)=\frac 13 P(2)\;\;\;\Rightarrow\;\;\;P(2)=3P(1).$$ In a similar way we see that $$P(3) = 7P(1)\;\;\mbox{and}\;\;\;P(4)=15P(1)$$
We are lead to conjecture that $$P(n) = \left(2^{n}-1\right)P(1)$$
And this is easily confirmed by induction.
To finish, we compute $P(1)$ from the other boundary value: $$P(1,000,002)=1=(2^{1,000,002}-1)P(1)$$
Whence $$P(1) = \frac {1}{(2^{1,000,002}-1)}$$
We now see the answer: $$P(1,000,000)= \frac{2^{1,000,000}-1}{2^{1,000,002}-1}$$
From which the desired inequality follows at once.
We note that $4$ times the numerator is $2^{1,000,002}-4$ so $4P(1,000,000)=1-\frac{3}{2^{1,000,002}-1}$. We may be below $\frac 14$ here, but not by much.
If the gambler ends up playing $\tau$ rounds, including the winning one, he has gained $36-\tau$ dollars and the sequence has a $(36/37)^{\tau-1}(1/37)$ chance of occurring. We have
$$E(W)=\sum_{k=1}^\infty(36/37)^{k-1}(1/37)(36-k)=-1$$
and
$$P(W>0)=P(\tau\le35)=1-P(\tau>35)=1-(36/37)^{35}$$
Now $\tau$ is a stopping time because the gambler stops when he has a positive net profit. Its expectation is infinite because $E(W)<0$ (cf. gambler's ruin).
Best Answer
Assuming that $T < \infty$ a.s., there is no way for $\mathbb{E}[X_T] > n$ if we enforce the no debt condition $X_m \ge 0$ for all $m$.
For any $m$ we have $\mathbb{E}[X_{T \wedge m}] = n$ by the optional stopping theorem (since $T \wedge m$ is a stopping time with finite expectation), and since $T < \infty$ a.s., $\lim_{m \rightarrow \infty} X_{T \wedge m} = X_T$ a.s. Since $X_{T \wedge m} \ge 0$ for all $m$, by Fatou's lemma we have \begin{align*} \mathbb{E}[X_T] = \mathbb{E}[\lim_{m \rightarrow \infty}X_{T \wedge m}] \le \liminf_{m \rightarrow \infty} \mathbb{E}[X_{T \wedge m}] \le n. \end{align*}
If we allow $\mathbb{P}(T = \infty) > 0$, it's unclear how to define $X_T$ on the event $\{T = \infty\}$.