After some consideration, in my opinion, "lower boundary" will make more sense rather than lower limit.
For example, this is the data,
Class Frequency
1 1
2 1
3 1
4 1
Based on the data, using we can know that the median is 2.5, without calculation. If using the formula as mentioned above, $\frac{n}{2}$ will get 2, there for the class contains the median is class 2, then using $L_m$ is a lower boundary,
$median = 1.5 + \left[ \frac{2 -1}{1}\right] \times 1 = 2.5$
This doesn't make sense for using lower limit. If changing the class to
Class Frequency
1-2 1
3-4 1
5-6 1
7-8 1
Using the method above, we will get,
$median = 2.5 + \left[ \frac{2 -1}{1}\right] \times 2 = 4.5$
However, if using class limit, then we will get 5.
This formula is the result of a linear interpolation, which identifies the median under the assumption that data are uniformly distributed within the median class.
To derive the formula, we can note that since $N/2$ is the number
of observations below the median, then $N/2 - F_{m-1}$ is the number of observations that are within the median class and that are below the median ($F_{m-1}$ is the cumulative frequency of the interval below the median class, i.e. of all classes below the median class).
As a result, the fraction $\displaystyle\frac {N/2 - F_{m-1}}{f_m}$ (where $f_m$ is the frequency of the median class) represents the proportion of data values in the median class that are below the median.
Now if we assume that data are uniformly distributed (i.e., equally spaced) within the median class, multiplying the last fraction by $c$ (total width of the median class) we obtain the fraction of median class width corresponding to the position of the median. Adding the result to $L_m$ (lower limit of the median class), we get the final formula $\displaystyle L_m + \left [ \frac { \frac{N}{2} - F_{m-1} }{f_m} \right ] \cdot c$, which identifies the median.
Best Answer
It is useful to identify the interval that contains the median, and that can be done without making unwarranted assumptions. For example, consider the $n = 25$ scores below, which have been sorted from smallest to largest.
Their exact median is $H = 102,$ the thirteenth observation in the sorted list.
A frequency histogram below based on the following cutpoints (bin boundaries):
These boundaries were chosen so that no (integer) score can fall exactly on a boundary.
The five interval midpoints are 79.5, 89.5, 99.5, 109.5, and 119.5. We can see from this frequency histogram below, that the corresponding frequencies are 4, 5, 6, 6, and 4.
Just looking at the histogram (or at a table of interval boundaries, midpoints, and frequencies), and without knowledge of the exact values of the $n = 25$ observations, all we can say about the median is that it falls in the interval $(94.5, 104.5)$ with midpoint $99.5$ and frequency $6.$ This interval is called the median interval.
In practice, grouped data tables and histograms are used mainly for samples that are at least moderately large. For a large sample it would ordinarily be sufficient to say that the median falls in the interval with midpoint $99.5.$
A favorite exercise in elementary statistics books is to try to approximate the exact value of the median from a histogram or from grouped data. Doing so requires one to make the assumption (seldom true) that the observations within the median interval are evenly spaced (or uniformly distributed).
One formula for approximating the exact median $H$ is
$$ H = L + \frac{w}{f_m}(.5n - cf_b),$$
where $L = 94.5$ is the lower limit of the median interval, $f_m = 6$ is the frequency of the median interval, $cf_b = 9$ is the number of observations in intervals below the median interval, $w = 10$ is the (common) interval width, and $n = 25$ is the total sample size. This kind of formula is sometimes called an 'interpolation' formula.
For our data,
$$ H = 94.5 + (10/6)(25/2 - 9) = 100.3333.$$
This procedure is seldom used in serious statistical analysis, and formulas for it can differ a bit from one textbook to another. I do not know the formula in your book, or why you wonder about a distinction between even and odd sample sizes $n$.
I hope this answer is helpful. If you are using a different formula to approximate the median, or if you have further questions, please leave me a Comment and edit your Question to be a little more specific. Then perhaps one of us can be of further help.
Notes: (1) The 25 observations are simulated from $\mathsf{Norm}(\mu = 100,\,\sigma = 15)$ and rounded to integers. So the median of the population from which the data were drawn is $\eta = 100.$ (2) It is not usually a good idea to 'group' datasets with $n$ as small as 25, or to make histograms of such small datasets. I chose this particular illustration because I thought it would make the application of the interpolation formula easy to follow.