**Edit** (Jan 28): As Didier points out in the comments, I made a mistake in my application of Chebyshev's inequality.

Didier and fedja already have gave you some great answers, but I'd like to go a little further. The reason all these arguments (including mine) are elementary is that $n$ is fixed, so it is tiny compared to $T$. Thus whether $n$ is $1$, finite, or even just growing very slowly compared to $T$, all the results will be qualitatively the same.

Suppose that the variables $X_i$ have finite variance $\sigma^2$, so that $$\mbox{$S(T)$ has mean $\mu T$ and standard deviation $\sigma \sqrt T$}.$$

In addition to the other answers controlling the expectation of the maximum, we can prove a stronger almost-sure result:

**Theorem.** Let $r > 1$ and $\epsilon > 0$. With probability one, there exists a (random) time $T_0$ so that for all $T \ge T_0$, $$ \max_{1 \le i \le n} S_i(T) \le \mu T + \epsilon T^{r/2}.$$

The proof is elementary, and only requires some basic theorems from probability (namely, Chebyshev's inequality and the Borel-Cantelli lemma).

Proof:

Then
$$ \mathbb P( \max S_i(T) \ge \mu T + \epsilon T^{r/2} ) = \mathbb P\left( \mbox{For some $i$, $S_i(T) \ge \mu T+ \epsilon T^{r/2}$} \right) $$
which equals $$\mathbb P\left( \bigcup_{i=1}^n \{S_i(T) \ge \mu T+ \epsilon T^{r/2}\} \right) \le n \cdot \mathbb P( S(T) \ge \mu T+ \epsilon T^{r/2})$$
by countable additivity. So it doesn't really matter that we're looking at the max of $n$ random variables or just a single one.

Now, let's analyze the right side of this expression using Chebyshev's inequality:
$$n \cdot \mathbb P( S(T) - \mu(T) \ge \sigma ( \tfrac{\epsilon}{\sigma} T^{r/2} ) ) \le \tfrac{\sigma^2}{\epsilon^2} \frac{n}{T^r}.$$

Since $r > 1$ and $n$ is fixed, the sum $\sum \tfrac{n}{T^r}$ is convergent, so the Borel-Cantelli lemma implies that, with probability one, the event $\{\max S_i(T) \ge \mu T + \sigma T^{r/2}\}$ occurs for finitely many values of $T$. This completes the proof.

**QED**

Some generalizations:

$n$ doesn't have to be finite. Suppose that $n = O(T^s)$. As long as $r - s > 1$, the series $\sum \tfrac{n(T)}{T^r}$ is still convergent, so the conclusion of the theorem still holds.

You can also modify the theorem above to the case that the variables $X_i$ have finite $(1+\epsilon)$th moment, for any $\epsilon$. Of course, the expression would change to $T^{r/(1+\epsilon)}$.

If you use the Central Limit Theorem, I believe that you can get the right side to be $\mu T + \sigma \sqrt T + O(\sqrt T)$. If you assume higher-than-second moment, the error term should be $o(\sqrt T)$.

The Law of the Iterated Logarithm gives an even better estimate. In your case, that should be: $$\mbox{With probability one, $\max_{1 \le i \le n} S_i(T) - \mu T \sim \sqrt{2 T \log \log T}$.}$$