There is a theorem (Erdos-Gallai) on degree sequences:
$\displaystyle d_i$ is a degree sequence of some graph if and only if
$\displaystyle \sum_{i=1}^{m} d_i \leq m(m-1) + \sum_{i=m+1}^{n} \min \{d_i, m\} \ \ \text{for} \ \ m \in \{1,2, \dots, n\}$
and
$\displaystyle \sum_{i=1}^{n} d_i$ is even.
It should be easy to verify for the case when $\displaystyle d_i = k \ \ \forall i \in \{1,2, \dots, n\}$ (cumbersome verification at the end of the answer).
The case $\displaystyle k=n-1$, we trivially know the existence of a regular graph ($\displaystyle K_n$).
Suppose $\displaystyle k \lt n-1$.
Now for $\displaystyle 1 \le m \le k$ we have that
$\displaystyle m(m-1) + (k-m)m + (n-k)k -mk = nk - k^2 - m \ge nk-k^2 - k = k(n-k-1) > 0$
For $\displaystyle m \gt k$ we have
$\displaystyle m(m-1) + (n-m)k - mk = m^2 - m(2k+1) + nk = (m-(2k+1)/2)^2 +nk - ((2k+1)/2)^2$
For $k \lt n-1$ we have that $\displaystyle 4nk \gt 4k^2 + 4k$
i.e. $4nk \ge 4k^2 + 4k + 1 = (2k+1)^2$.
Hence
$\displaystyle m(m-1) + (n-m)k - mk \ge 0$
Thus if $\displaystyle kn$ is even, there exists a $k$-regular graph on n vertices.
The other part I leave to you.