The problem asked (without independence) can be solved. Fix nonnegative real numbers $a_1, …, a_n$. Define $M = \sum_{i=1}^n a_i$ and assume $M \geq 1$. Define $$\mathcal{X} = \{0, a_1\} \times \{0, a_2\} \times … \times \{0, a_n\} $$ To solve the problem, we first solve a simpler problem (called "Problem 1") below.
Problem 1: Fix $c \in [1,M]$. We want to find a random vector $(X_1, ..., X_n)$ to solve \begin{align} \mbox{Minimize:} \quad & P\left[\sum_{i=1}^n X_i \geq c\right] \\ \mbox{Subject to:} \quad & E\left[\sum_{i=1}^n X_i\right] = c \\ & (X_1, ..., X_n) \in \mathcal{X} \end{align}
To solve Probelm 1, define $$ g_c = \max_{(x_1, ..., x_n) \in \mathcal{X}} \left\{\sum_{i=1}^n x_i : \sum_{i=1}^n x_i < c\right\} $$ That is, $g_c$ is the largest value of $\sum_{i=1}^n x_i$ over all vectors $(x_1, ..., x_n) \in \mathcal{X}$ with components that sum to less than $c$. The max is achievable because $\mathcal{X}$ is a finite set that contains the all-zero vector. Note that $$0 \leq g_c < c \leq M$$ Define $\vec{x}_c$ as an element of $\mathcal{X}$ with a sum of components that is equal to $g_c$. Define $$p_c = \frac{c-g_c}{M-g_c}$$ Observe that $0< p_c \leq 1$. Define the random vector $$ (Z_1, ..., Z_n) = \left\{ \begin{array}{ll} (a_1, ..., a_n) &\mbox{ with prob $p_c$} \\ \vec{x}_c & \mbox{ with prob $1-p_c$} \end{array} \right.$$ It is not difficult to show that $E[\sum_{i=1}^nZ_i] = c$ and $$P\left[\sum_{i=1}^n Z_i \geq c\right] = p_c $$
Claim
The random vector $(Z_1, ..., Z_n)$ solves problem 1. In particular, the minimum probability is: $$ P\left[\sum_{i=1}^n Z_i \geq c\right] = p_c = \frac{c-g_c}{M-g_c}$$
Proof
Define $$\mathcal{S} = \left\{\sum_{i=1}^n x_i \in \mathbb{R} : (x_1, …, x_n) \in \mathcal{X}\right\}$$ Let $(X_1, ..., X_n)$ be any random vector in $\mathcal{X}$ that satisfies $E[\sum_{i=1}^n X_i]=c$. Define $S=\sum_{i=1}^n X_i$. Note that: $$ S \in \mathcal{S} \subseteq [0, g_c] \cup [c, M]$$ Thus \begin{align} c &= E[S|S\geq c]P[S\geq c] + E[S|S\leq g_c](1-P[S\geq c]) \\ &\leq M P[S\geq c] + g_c(1-P[S\geq c]) \end{align} and so $$ P[S\geq c] \geq \frac{c-g_c}{M-g_c} = p_c $$ Thus, if $(X_1, ..., X_n)$ is any vector in $\mathcal{X}$ that satisfies the constraints of Problem 1 (namely, $E[\sum_{i=1}^n X_i]=c$), then $$ P\left[\sum_{i=1}^n X_i \geq c\right] \geq p_c $$ On the other hand, the lower bound is achieved by the random vector $(Z_1, ..., Z_n)$ defined above. $\Box$
Problem 2: Find a random vector $(X_1, ..., X_n)$ to solve \begin{align} \mbox{Minimize:} \quad & P\left[\sum_{i=1}^n X_i \geq E\left[\sum_{i=1}^nX_i\right]\right] \\ \mbox{Subject to:} \quad & E\left[\sum_{i=1}^n X_i\right] \geq 1 \\ & (X_1, ..., X_n) \in \mathcal{X} \end{align}
Problem 2 can be solved via Problem 1 by optimizing over the mean $c = E[\sum_{i=1}^n X_i]$ over all $c \in [1, M]$. Thus:
\begin{align} p^* &= \inf_{c \in [1, M]} \frac{c-g_c}{M-g_c} \end{align}
This can be solved by considering two cases:
Case 1: Suppose there is an $x \in \mathcal{S}$ such that $1\leq x < M$. Then we can choose $c = x + \epsilon$ for some very small value $\epsilon>0$, so that $g_c=x$, and $(c-g_c)/(M-g_c)$ is arbitrarily small. In this case, $p^*=0$.
Case 2: Suppose there is no $x \in \mathcal{S}$ such that $1\leq x < M$. So $g_c=g_1$ for all $c \in [1,M]$. So we choose $c=1$ and $$ p^* = \frac{1-g_1}{M-g_1}$$