No $\kappa>1$ implies any bound on $d$. Without loss of generality, let $2\geq\kappa>1$. Assume that we have already found a $\Lambda$ of cardinality $d-1$ satisfying $\mu(\Lambda)\leq\kappa$. We shall construct a $\Lambda'$ of cardinality $d$ satisfying $\mu(\Lambda')\leq\kappa$.

Let $\lambda_1,\dots,\lambda_{d-1}$ be the elements of $\Lambda$, and let $\sigma$ be their sum. Choose any positive integer $\lambda_d$ divisible by all positive integers up to $\sigma/(\kappa-1)$. We claim that $\Lambda':=\Lambda\cup\{\lambda_d\}$ satisfies $\mu(\Lambda')\leq\kappa$, i.e. for any integer $l\geq 2$, we have
$$ \sum_{i=1}^d (\lambda_i\ \ \text{mod}\ \ l)\leq\kappa l.$$
We distinguish between two cases. If $l\leq\sigma/(\kappa-1)$, then the last term $(\lambda_d\ \ \text{mod}\ \ l)$ vanishes, hence the required inequality follows from the initial assumption
$$ \sum_{i=1}^{d-1} (\lambda_i\ \ \text{mod}\ \ l)\leq\kappa l.$$
If $l>\sigma/(\kappa-1)$, then $(\lambda_i\ \ \text{mod}\ \ l)=\lambda_i$ for $1\leq i\leq d-1$, while $(\lambda_d\ \ \text{mod}\ \ l)<l$, hence the required inequality follows from $\sigma+l<\kappa l$.