Odd order moments of a symmetrical distribution

I'm going to assume you mean (1) symmetric about $0$, and (2) distributions for which the odd-order moments actually exist. Under these assumptions, yes.

In the continuous case, the $n$th moment for a distribution with pdf $f(x)$ exists and is $$\int_{-\infty}^{\infty} x^n f(x) \,dx,$$ exactly when the integral converges. So, what does it mean for the integral to converge?

Remember that we define $\int_{-\infty}^{\infty} x^n f(x) \,dx$ to be $$ \int_{-\infty}^0 x^n f(x) \, dx + \int_0^{\infty} x^n f(x) \, dx.$$ (Well, we could choose some point other than $0$ to split the integral, but $0$ works fine.) Convergence of $\int_{-\infty}^{\infty} x^n f(x) \,dx$ is equivalent to the convergence of both of these integrals. Existence of the $n$th odd-order moment, then, is equivalent to $\int_{-\infty}^0 x^n f(x) \, dx$ and $\int_0^{\infty} x^n f(x) \, dx$ both being finite.

Suppose $\int_0^{\infty} x^n f(x) \, dx = A$. If we let $u = -x$, then $$\int_{-\infty}^0 x^n f(x) \, dx = \int_{\infty}^0 (-u)^n f(-u) \, (-du) = (-1)^n \int_0^{\infty} u^n f(-u) \, du = - \int_0^{\infty} u^n f(u) \, du = -A,$$ where the first step is legitimate because the integral converges and the third step follows because $n$ is odd and $f$ is symmetric about the origin.

Thus, if the $n$th odd-order moment exists, it is equal to $\int_{-\infty}^{\infty} x^n f(x) \,dx = -A + A = 0.$

The discrete case for $n$ odd is analogous.

I think it's easier than arguments about integrals and sums. Saying a random variable $X$ has a symmetric distribution means that $X$ and $-X$ have the same distribution. This means that for any function $f$, if $E[f(X)]$ exists then so does $E[f(-X)]$ and they are equal. So take $f(x) = x^n$ with $n$ odd. If $X$ has an $n$th moment, then $E[X^n]$ exists and so $$E[X^n] = E[(-X)^n] = E[(-1)^n X^n] = E[-(X^n)] = -E[X^n]$$ which means $E[X^n] = 0$.