Assume that $Y_1, Y_2, ..., Y_{100}$, are the rounding errors made on one hundred items and that these are independent and uniformly distributed over the interval $[−50(\text{USD}), 50(\text{USD})]$.
By the distribution assumption made for each $Y_i$, $E(Y_i)=0$ for $i =1,2,...,100$
Why is $E(Y_i)=0$?
On
The values of the underlying amount variables $X_i$ may come from the most diverse sources: from items sold at integer dollar prices, from sales with 5ct or even 1 ct precision, or from currency conversions. It follows that the $X_i$ are for all purposes continuous real variables. Our investigator is inclined to throw the small change away, i.e., he will only deal with hundreds of dolloars, obtained by rounding. The rounding errors $Y_i$ in the question are then continuous real variables in the range $[-50,\>50]$. Uniformly distributed means that for any subinterval $[a,b]\subset[-50,\>50]$ the probability $P\bigl[Y_i\in[a,b]\bigr]$ is equal to the ratio ${b-a\over100}$, and independently means that even if you are told the true values of the first ninetynine $Y_i$, this gives you no hint whatsoever what the value of $Y_{100}$ could be. Whether these assumptions are justified in the case at hand is not discussed; they just form the base of the mathematical model.
Simple symmetry considerations should make it intuitively clear that $E(Y_i)=0$ in such a model.
No, not quite. What you have provided in the comments is the density. For simplicity, $Y_i = X$, and what you have given is $$f_X(x) = \frac{1}{b-a} = \frac{1}{100}.$$
Then to find the expectation, we follow the usual formula $$E[X] =\int_{-50}^{50} x\cdot f_X(x)\,dx = \int_{-50}^{50}x\cdot\frac{1}{100}\,dx = 0.$$
Somewhere along the way, you were probably given that for a random variable $X$, if it follows $\text{unif}[a,b]$, then $$E[X] = \frac{a+b}{2}.$$
You can verify this. It also gives the same result, $E[X] = \frac{-50+50}{2} = 0$.
Addendum:
In the case that it follows a discrete uniform distribution, $P(X = k)$ for some $k$ in $[-50, \dotsc,50]$ is $\frac{1}{101}$. So using the usual formula for discrete cases, we have $$\sum_{k = -50}^{50} kP(X=k) = 0.$$
You can verify that $E[X] = \frac{a+b}{2}$ for the discrete case too.