Let us try to find solutions for smaller versions of the problem before we generalise to $16$ players.
For example, if we begin with just $4$ players and are asked to make consecutive room assignments based on the same criteria, how would we achieve it?
It turns out, that in this case, as long as one of the odd-numbered rooms in the sequence is divisible by $3$, we can achieve a valid room assignment.
For example, we can assign players $(1,2,3,4)$ to rooms $(7,6,9,8)$.
Another way to put this is that we can assign players $(1,2,3,4)$, in some order, to rooms $\big(6n+k, 6n+k+1, 6n+k+2, 6n+k+3\big)$ for any positive integer $n$, for any $k$ in the set $\{ 0, 1, 2, 3\}$
Now, let us consider the scenario where we have $8$ players.
We already know from the previous discussion that we can assign the players $2,4,6$ and $8$ simply by doubling the classes of solutions found previously. That is $(2,4,6,8)$ can be assigned to $\big(12n+2k, 12n+2k+2, 12n+2k+4, 12n+2k+6\big)$ for any positive integer $n$ and for any $k$ in the set $\{ 0, 1, 2, 3\}$. Then, we can search for the sub-cases of $n$ and $k$ for which we can fill in the odd numbers in between. Notice that, for fixed $k$, there will be a repeating pattern after every $35$ values of $n$. That is, a solution for $(n,k)$ implies a solution for $(n+35N, k)$ with $N$ a positive integer.
We can break down our analysis to look at the values of $k$ separately.
k=0
In this case, the only odd number divisible by $3$ in the range will be $12n+3$, so that is where player $3$ must be placed. This means that within the set $\{12n-1, 12n+1, 12n+5, 12n+7\}$ one must be divisible by $5$ and a different one by $7$ but the assignments cannot be made to both the first and last entries. It's easily seen that the number of ways to do this is $2\left(\binom{4}{2} - 1\right) = 10$.
The analysis is a little tedious here but can be done by hand. The $10$ values of $n$ for which this happens are $\{4, 7, 10, 13, 17, 18, 25, 27, 34, 35\}$. In particular, for any $m \geq 0$, the solutions for the $8$-player case within this class are
those rooms in the range $[420m+12n-1,\, 420m+12n+6]$ for $n \in \{ 10, 13, 17, 18, 25, 27\}$ and
those rooms in the range $[420m+12n, \,420m+12n+7]$ for $n \in \{ 4, 7, 25, 27, 34, 35\}$
k=1
The analysis in this case is very similar (there is a slight additional subtlety as $12n+9$ is also in this range which gives us access to a few more solutions), so I won't go into all the details but for any $m \geq 0$, the solutions for the $8$-player case within this class are
those rooms in the range $[420m+12n+1,\, 420m+12n+8]$ for $n \in \{ 4, 7, 25, 27, 34, 35\}$ and
those rooms in the range $[420m+12n+2, \,420m+12n+9]$ for $n \in \{ 5, 6, 13, 15, 19, 21, 28, 29, 34, 35\}$
k=2
For any $m \geq 0$, the solutions for the $8$-player case within this class are
those rooms in the range $[420m+12n+3,\, 420m+12n+10]$ for $n \in \{ 5, 6, 13, 15, 19, 21, 28, 29, 34, 35 \}$ and
those rooms in the range $[420m+12n+4, \,420m+12n+11]$ for $n \in \{ 7, 9, 27, 30, 34, 35\}$
k=3
For any $m \geq 0$, the solutions for the $8$-player case within this class are
those rooms in the range $[420m+12n+5,\, 420m+12n+12]$ for $n \in \{ 7, 9, 27, 30, 34, 35 \}$ and
those rooms in the range $[420m+12n+6, \,420m+12n+13]$ for $n \in \{ 7, 9, 16, 17, 21, 24 \}$
Now let us consider the case of $16$ players.
We already know from the previous discussion that we can assign the even-numbered players simply by doubling the classes of solutions found in the previous paragraph. Then, we can search for the sub-cases of $m$, $n$ and $k$ for which we can fill in the odd numbers in between.
A useful thing to note here is that we will be trying to assign $5$ and $15$ to odd-numbered rooms so, in our run of $16$ consecutive room numbers, there must be two which end in $5$. This means that the last digit of the first element of the consecutive room numbers must be either $0, 1, 2, 3, 4$ or $5$.
This leaves us with the following number ranges to test:
$[840m+24n-3,\, 840m+24n+12]$ for $n \in \{ 17, 27\}$,
$[840m+24n-2,\, 840m+24n+13]$ for $n \in \{ 13, 18\}$,
$[840m+24n-1, \, 840m+24n+14]$ for $n \in \{ 4, 34\}$,
$[840m+24n, \, 840m+24n+15]$ for $n \in \{ 25, 35\}$,
$[840m+24n+1,\, 840m+24n+16]$ for $n \in \{ 25, 35\}$,
$[840m+24n+2,\, 840m+24n+17]$ for $n \in \{ 7, 25, 27, 35\}$,
$[840m+24n+3, \,840m+24n+18]$ for $n \in \{ 5, 13, 15, 28, 35\}$,
$[840m+24n+4, \,840m+24n+19]$ for $n \in \{ 5, 15, 19, 29, 34, 35\}$,
$[840m+24n+5,\, 840m+24n+20]$ for $n \in \{ 5, 15, 19, 29, 34, 35 \}$,
$[840m+24n+6,\, 840m+24n+21]$ for $n \in \{ 6, 19, 21, 29, 34 \}$,
$[840m+24n+7, \,840m+24n+22]$ for $n \in \{ 7, 9, 27, 34\}$,
$[840m+24n+8, \,840m+24n+23]$ for $n \in \{ 9, 34\}$,
$[840m+24n+9,\, 840m+24n+24]$ for $n \in \{ 9, 34\}$,
$[840m+24n+10,\, 840m+24n+25]$ for $n \in \{30, 35\}$,
$[840m+24n+11, \,840m+24n+26]$ for $n \in \{16, 21 \}$,
$[840m+24n+12, \,840m+24n+27]$ for $n \in \{7, 17 \}$,
For each $m$, this is $50$ cases to check but we can start with $m=0$
In each case, it's reasonably quick to check by hand (often the cases for $13$ and $11$ can be ruled out pretty quickly). For example, $n = 34$ and $35$ appear quite frequently so we can test for primality in the range following these numbers. It turns out that, in the case $m=0$, there is no solution (although the range beginning at $363$ comes very close).
Next, we try $m=1$. Again, testing through the cases, I've found no solutions in this range.
Next, we try $m=2$. This time there are two cases which work!
These are $(m,n,k) = (2,34,8)$ and $(2,34,9)$