Lets do the algebra, let $X_{ji}, j=1,\dotsc,k ; i=1,\dotsc,n$ be independent observations from $k$ groups, each with expectation $\mu_j, j=1,\dotsc,k$ and variance $\sigma^2$.
The group means is $\bar{X}_j= \frac1{n}\sum_i X_{ji}$ and the total mean is $\bar{\bar{X}} = \frac1{nk} \sum_j \sum_i X_{ji} = \frac1{k} \sum_j \bar{X}_j$ with variances $\sigma^2/n$ and $\sigma^2/nk$, respectively. We calculate (for $j\not = l$):
$$ \DeclareMathOperator{\var}{\mathbb{V}ar}
\DeclareMathOperator{\cov}{\mathbb{C}ov}
\DeclareMathOperator{\corr}{\mathbb{c}or}
\cov(\bar{X}_j-\bar{\bar{X}},\bar{X}_l-\bar{\bar{X}}) =
\cov(\bar{X}_j,\bar{X}_l) - \cov(\bar{X}_j,\bar{\bar{X}})-\cov(\bar{X}_l,\bar{\bar{X}}) + \cov(\bar{\bar{X}},\bar{\bar{X}})
$$
Each of the terms:
$$
\cov(\bar{X}_j,\bar{X}_l) = 0 \quad (j\not= l) \\
\cov(\bar{X}_j,\bar{\bar{X}}) = \cov(\bar{X}_l,\bar{\bar{X}}) = \\\cov(\bar{X}_j, \frac1k(\bar{X}_1+\dotso+\bar{X}_j+\dotso+\bar{X}_k)) =
\cov(\bar{X}_j,\frac1k \bar{X}_j)= \frac1k \cov(\bar{X}_j,\bar{X}_j) =
\frac1k \var(\bar{X}_j )= \sigma^2/nk \\
\cov( \bar{\bar{X}}, \bar{\bar{X}})=\var(\bar{\bar{X}})=\sigma^2/nk
$$
Putting everything together, we have
$$
\cov(\bar{X}_j-\bar{\bar{X}},\bar{X}_l-\bar{\bar{X}}) = 0 - \sigma^2/nk - \sigma^2/nk + \sigma^2/nk = -\sigma^2/nk
$$
and finally the correlation is
$$
\corr(\bar{X}_j-\bar{\bar{X}},\bar{X}_l-\bar{\bar{X}}) =
\frac{-\sigma^2/nk}{\sigma^2/n}= -\frac1k
$$
and since this correlation is the same for each pair $(j,l) \quad j\not= l$ we say it is equicorrelated. Your source above gave $k-1$ not $k$ in the denominator, you should check if you cited correctly, because what we have here is correct. You can check that intuitively by looking on the case $k=2$, your formula gives then a correlation $-1$ which cannot be correct.
