I suggest identifying each group with a grouping variable. groupID is the grouping variable.
% Set these values
n = 540; % number of data points
nPerGroup = 20; % number of consecutive values per group
% compute group IDs
nGroups = n/nPerGroup; % number of groups (must be an integer)
assert(mod(nGroups,1)==0, 'nGroups must be an integer') % assumption check
groupID = repelem((1:nGroups)', nPerGroup, 1); % group ID (same length as your data (540 elements)
For 540 values and 20 values per group, there are 27 partitions for 27-fold cross validation. Note, however, that k-fold cross validation uses random samples and you are not sampling randomly unless the list of 540 values are already randomized so if these methods are described in a publication, you'd need to inditate this adjustment to the methodology. For each repetition one partition is reserved for testing while the other k-1 partitions are used for training. This also departs from your explanation that one partition is used for training and another for testing.
To implement 27-fold cross validation, you can loop through each groupID as follows, data is your 540-element data which is assumed here to be a vector
for i = 1:nGroups
train = data(groupID ~= i);
test = data(groupID == i);
% < do Training >
% < do testing >
end
