If $(R,m)$ is a local Noetherian reduced ring of Krull dimension $1$ then $R$ is Cohen-Macaulay, since in a reduced Noetherian ring the set of zero divisors is the (finite) union $U$ of minimal prime ideals, so there exists an element $x\in m$ which is not a zero divisor (otherwise, $m$ lies in $U$ and equals to one of the minimal prime ideals by prime avoidance, which is a contradiction since the height of $m$ is $1$). Now, the singleton $\{x\}$ would be an $R$-sequence, and since the grade of $m$ is at most equal to the height of $m$ the equality occurs.
Now, if $R$ is not local what is the proof?
Thanks for any cooperation!
Let $R$ be a noetherian reduced ring, $\dim R=1$, and $m$ a maximal ideal of $R$. Then $R_m$ is noetherian, reduced, and $\dim R_m\le 1$. If $\dim R_m=1$ you are in the local case, so $R_m$ is CM. If $\dim R_m=0$ there is nothing to prove since every zero-dimensional noetherian ring is CM.