Solutions to the equation $y^Tx\geq L$ for $y\in \mathbb{R}^k$ is a half space $H$ with the vector $Lx/(x^Tx)$ lying on the boundary, with the interior of the half space in the normal direction $x$ from there. A difficulty in this situation is how the boundary of the half space sits with respect to the integer lattice... since you are allowing $x$ to have real coefficients, it may potentially avoid all integer points!
One approach would be to approximate $x$ by a vector of $\mathbb{Q}^k$, giving an actual integer linear program once one clears out the denominators. Either you can do this for a sequence of better approximations and stop once the minimal solution seems good enough, or it might be possible to calculate how good of a solution one would need to guarantee the approximated half space contains exactly the same points of $\mathbb{N}^k$. I have not read it, but this thesis appears to have quite a lot about these sorts of approximations: Vaughan Clarkson, "Approximation of Linear Forms by Lattice Points". One keyword appears to be "simultaneous Diophantine approximation."
There is a chance that your $x$ is already in $\mathbb{Q}^k$, for instance if you are on a computer with a floating-point representation of the vector, in which case approximation might not be necessary.
I'm going to leave here my own attempt at a solution. First of all, the $L=0$ case has the minimal solution $n=0$, so we may assume $L>0$ and then divide $x$ through by $L$ to simplify the problem to $y^Tx\geq 1$. Let $f(n)=\sum_{i=1}^kn_i$ be your objective function, which of course is $f(n)=\mathbf{1}^Tn$ with $\mathbf{1}$ the all-ones vector.
The next observation is that we can assume $x_i>0$ for each $i$. If not, then by taking any solution to $n^Tx\geq 1$ we can get another solution by setting $n_i$ to $0$, possibly decreasing $f(n)$ in the process, and then removing dimension $i$ from the problem.
Suppose $e\in \mathbb{R}^k_{\geq 0}$ is a vector of non-negative entries. The inequality $n^T(x+e)\geq 1$ has all the solutions to $n^Tx\geq 1$ plus possibly some more. Given an integer $N\geq 1$, we can form the rational approximation $x^N\in\mathbb{Q}^k_{>0}$ given by $x^N_i=\lceil Nx_i\rceil/N\geq x_i$, hence $n^Tx^N\geq 1$ has all the solutions to $n^Tx\geq 1$.
Let $i$ be such that $x_i$ is largest, and let $M=\lceil 1/x_i\rceil$. We can see there is an $f$-minimal solution within the set $\{0,1,2,\dots,M\}^k$, since $n$ given by $n_i=M$ and $n_j=0$ for $j\neq i$ is a solution with $f(n)=M$. So, there are at most $(M+1)^k$ solutions to consider. Finiteness implies that there is some large enough $N$ such that $n^Tx^N\geq 1$ has the same solutions as $n^Tx\geq 1$, when restricted to $n\in \{0,1,2,\dots,M\}^k$. In particular, the sets $S_N\subseteq \{0,1,2,\dots,M\}^k$ of solutions to $n^Tx^N\geq 1$ satisfy $S_1\supseteq S_2\supseteq S_3\supseteq\cdots$, so it must eventually stabilize.
Since $x^N_i-x_i<\frac{1}{N}$, we have for a minimal solution $n$ that
\begin{align}
n^Tx^N &= \sum_{i=1}^k n_i x_i^N\\
&< \sum_{i=1}^k n_i (x_i+\frac{1}{N}) \\
&= \sum_{i=1}^k n_i x_i + \frac{1}{N}\sum_{i=1}^k n_i \\
&= n^Tx + \frac{1}{N} f(n)\\
&\leq n^Tx + \frac{M}{N}.
\end{align}
So $n^Tx^N\geq 1$ implies $n^Tx\geq 1-\frac{M}{N}$. This new inequality defines a half space $H_N\supset H$; letting $S'_N=H_N\cap\{0,1,2,\dots,M\}^k$, we have
$\require{AMScd}$
\begin{CD}
S_1 @>\supseteq>> S_2 @>\supseteq>>\cdots \\
@VV\subseteq V @VV\subseteq V \\
S_1' @>\supseteq>> S_2' @>\supseteq>>\cdots
\end{CD}
If we can estimate some $N$ such that $S'_N = H\cap \{0,1,2,\dots,M\}^k$, which is a point from which the bottom sequence stabilizes, then the top sequence has stabilized as well. (Geometrically, inside of $\mathbb{R}^k_{\geq 0}$, we have that the boundary of rational approximation half space lies between the boundaries of $H$ and $H_N$.)
This is where I got stuck: we need to know some $\epsilon>0$ distance we can translate $H$ in the $-x$ direction without its boundary colliding with the integer lattice within $\mathbb{R}^k_{\geq 0}$. Once we have such an $\epsilon$, we can calculate a sufficient $N$ and then do integer programming with the corresponding rational approximation of the inequality.
There is still an algorithm here, however. Choose an $N$, do the corresponding integer program to get a preliminary solution $n'$. If $n'$ does not satisfy $(n')^Tx\geq 1$, then increase $N$ to a value where $n'$ does not satisfy $(n')^Tx^N\geq 1$ and repeat. Eventually $n'$ will be a solution to the unapproximated problem. Since $n'$ is $f$-minimal for $n^Tx^N\geq 1$, it is $f$-minimal for $n^Tx\geq 1$ as well. Without the $\epsilon$ estimate, though, the algorithm is merely guaranteed to terminate.
By the way, it is conceivable that there is a method that could use less-good approximations of the inequality with a smaller $N$ by somehow taking more advantage of the particular objective function.
