This problem is NP-hard by a reduction from partition.
Let $X = \{x_1, \dots, x_n\}$, be an instance of partition and let $2M = \sum_{x \in X} x$ (assume that $M$ is an integer as otherwise the instance is trivial).
Assume that each $x_i$ is positive and let $\ell = \lceil \log 2M \rceil$.
I will now construct some values based on $X$. I will denote a generic value by a tuple $v$ of $n+2$ entries $(v^{*}, v^{(0)}, v^{(1)}, \dots, v^{(n)})$, where all entries $v^{(i)}$, $i=0,\dots,n$ are between $0$ and $2^\ell - 1$.
The tuple $v$ will actually correspond to integer whose binary representation is obtained by encoding $v^{*}$ in binary (using a variable number of bits), each of $v^{(0)}, \dots, v^{(n)}$ using $\ell$-bits, and finally concatenating the resulting bitstrings, in order.
An example of the tuple $v=(5, 2, 7, 9)$ with $\ell = 4$ is as follows:
$$
(\underbrace{101}_{v^*}
\underbrace{0010}_{v^{(0)}}
\underbrace{0111}_{v^{(1)}}
\underbrace{1001}_{v^{(2)}})_2 = (21113)_{10}
$$
Notice that the value obtained in this way has an encoding of polynomial length w.r.t. the length of the encoding of the original partition instance, as long as $v^{(*)}$ is small enough (as will be the case). The sum of two tuples is the tuple whose integer value equals the sum of the integer values of the summands.
Let then $y_i = (1, x_i, 0, \dots, 0, 1, 0, \dots, 0)$, where the second entry set to $1$ is $y_i^{(i)}$.
Similarly, let $z_i = (1, 0, 0, \dots, 0,1,0, \dots, 0)$, where $z_i^{(i)} = 1$.
The instance of the problem in the question consists of the set of numbers $S = \{y_i : x_i \in X\} \cup \{z_i : x_i \in X\}$ and of the target value $T=(n,M,1, \dots, 1)$.
Given a solution $X' \subseteq X$ to the partition problem, it is easy to construct a solution $S' \subseteq S$ for our problem such that $\sum_{x \in S'} x = T$. Simply select $S' = \{ y_i : x_i \in X'\} \cup \{ z_i : x_i \in X \setminus X' \}$ (notice that it suffices for $S'$ to be a set rather than a multi-set).
I will now argue that if a (multi-)set $S' \subseteq S$ such that $\sum_{x \in S'} x = T$ exists, then the partition instance $X$ has answer "yes".
From now on let $S'$ be such a set and notice that, by our choice of $T$, the overall multiplicity of all the elements of $S'$ cannot be larger than $n$ (as otherwise $T^{(*)}$ would be exceeded).
Claim: Let $m(x)$ be multiplicity in $S'$ of $x \in S$.
$\forall i=1,\dots,n$ we have $m(y_i) + m(z_i) = 1$.
Proof:
If $m(y_i) + m(z_i) = 0$ then, since $T^{(i)}=1$, $S'$ must contain at least $2^\ell \ge 2M > n$ elements $y_j$ or $z_j$ such that $j > i$, contradicting $\sum_{x \in S'} m(x) \le n$.
Assume then that $m(y_i) + m(z_i) > 1$, and consider the tuple $h$ corresponding to the sum of exactly 2 copies of $y_i$ and/or $z_i$ in $S'$.
We have $h^{(i)}=2$ and, since $T^{(i)}=1$, we conclude that there must be at least $2^\ell - 2$ other copies of some $y_j$ or $z_j$ for $j \ge i$ in $S'$.
The total multiplicity of the elements in $S'$ is therefore at least $2 + 2^\ell - 2 > n$, a contradiction. $\;\; \square$
Let $X' = \{x_i : y_i \in S' \}$ and call $h = \sum_{x \in S'} x$. Since there can be at most one copy of each $z_i$ in $S'$ (see the above Claim), the only way to affect $h^{(0)}$ is through the values $y_i \in S'$. Since there can be at most one copy of each $y_i$ in $S'$ (see the above Claim), we must have $\sum_{x_i \in X'} x_i = \sum_{y_i \in S'} y_i^{(0)} = h^{(0)} = T^{(0)} = M$.
