Problem about expectation of maximum partial sum Given a number $m$, a random composition (strong) of this number into $n$ positive parts so that we can get $n$ random variable $X_1, X_2,\dots, X_n$ with $$X_1+X_2+\cdots+X_n=m$$ 
Note that all compositions of $m$ have the same probability.
Let $Y_i = X_i - \mathbb{E}(X_i)$, $S_i = Y_1 + \cdots + Y_i$
I want to calculate $\mathbb{E}(\max_{1 \le i \le n} S_i)$.
 A: If I understand the problem correctly then $\mathbb{E}(X_i)=\frac{m}{n}$  So that the $Y_i$ are values from $0-\frac mn,1-\frac{m}{n},\dots,m-\frac mn$ which add to $0$. 
If we would instead compose $m-n$ as an ordered sum of $n$ non-negative integers this would have the effect of lowering the $X_i$ and expectation by one each but would give the same $Y_i$ and expectation for the maximum of the $S_i.$ That way of setting up the problem might lead to neater expressions.
It seems to me that the expected maximum would asymptotically be $$c_nm+e_r+O(1/m)$$ where $e_r$ depends on the congruence class of $m$ mod $n.$  The fuzzy reasoning is that if we were to take a composition of $m$ and triple all the entries we would get a composition of $m'=3m$ with thrice the expected maximum. To get an arbitrary composition of $ 3m$ we might move some of the parts up or down by 1, but that would not have a big effect on the ratio of the expected maximum to $m'$. Beyond that I'll just say that the calculations below support this supposition with $$c_2=\frac{1}{8},c_3=\frac{4}{27},c_4=\frac{39}{256},c_5=\frac{472}{3125}.$$ 
More specifically, calculations strongly suggest for each $0 \le r \le n-1$ there is a  polynomial $p_r(q)$ of degree $n$ such that for $m=qn+r$ $$\mathbb{E}(S_{\max})=\frac{p_r(q)}{(m-1)(m-2)\dots(m-n)}.$$ In all cases the leading coefficient is the same, namely $n^{n}c_n.$ Assuming that that is the case, one can find the coefficients of $p_r(q)$ and hence $c_n$ by computing $\mathbb{E}(S_{\max})$ for $m=r,n+r,2n+r,\dots,n^2+r.$ And going a bit further provides a check. Eventually this would be too hard a calculation, but the obvious naive strategy works well for a while (I did $n \le 4$ and $m \le 70$ with Maple.)
later  N. Elkies' great answer explains (and establishes) these facts and much more, but I leave this amended naive approach for what it is worth.
For $n=2$  the expected value of the maximum is $\frac{m-1}{8}$ or $\frac{m-1}{8}-\frac{1}{8(m-1)}$ according as $m$ is odd or even. In the odd case, for $1 \le X_1 \le \frac{m-1}{2}$ the maximum is $S_2=0.$ And for $\frac{m+1}{2} \le X_1 \le m-1$ the maximum is $S_1=X_1-\frac{m}{2}$ with expected value $\frac{1/2+(m-2)/2}{2}=\frac{m-1}{4}.$ When $m$ is even there is also the composition into 2 equal parts with $S_1=S_2=0$ and it slightly lowers the answer. After some arithmetic the expected maximum turns out to be as stated.
For $n=3$ the expected maximum appears to be about $\frac{4}{27}m$ and exactly 
$\frac{2(q-1)^2(2q-3)}{(m-1)(m-2)},\frac{2(q-1)q(2q-3)}{(m-1)(m-2)}$ or $\frac{2q^2(2q-3)}{(m-1)(m-2)}$ according as $m=3q-1,m=3q$ or $m=3q+1.$ The sequence A200067 from the OEIS may be related.
In the case that $n=4$ the expected maximum appears to be  $\frac{39}{256}m+O(1)$ In the case $m=4q$ the expected maximum appears to be exactly $\frac{3(13q^2-13q+3)q(q-1)}{(m-1)(m-2)(m-3)}.$ When $m=4q-1$ or $m=4q+1$ the term q(q-1) is replaced by $(q-1)^2$ or $q^2$ respectively. This is similar to the $n=3$ case above. For $m=4q+2$ it comes out to be $\frac{3q^2(13q^2+2)}{(m-1)(m-2)(m-3)}$
For $n=5$ similar considerations give an expected maximum of $\frac{472}{3125}m+O(1)$
A: I can't comment , but based on Noam D. Elkies's numbers and OEIS A000435 it looks as if 
$$c_n = \frac{(n-1)!}{2 n^n} \sum_{k=0}^{n-2} \frac{n^k}{k!}.$$
