If $P$ is on $AB$, you can take $Q = A$ and $R$ on $OC$ such that $\overrightarrow{OR} = \overrightarrow{AP}$; if $P$ is on $BC$, you can take $Q = C$ and $R$ on $OA$ such that $\overrightarrow{OR} = \overrightarrow{CP}$; if $P$ is on $CA$, you can take $Q = P$, and $R = O$. If the point $P$ is not on the boundary of $\triangle ABC$, then $\overrightarrow{OP} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OC}$ with $\alpha$ and $\beta$ being rational numbers satisfying $0 < \alpha, \beta < 1$. Let the least common denominator of $\alpha$ and $\beta$ be $n$. For each integer $k$, let $\alpha_k = \{ k\alpha \}$ and $\beta_k = \{ k\beta \}$. Therefore, $(\alpha_k, \beta_k) \neq (0, 0)$ when $1 \leq k \leq n - 1$ and $(\alpha_n, \beta_n) = (0, 0)$. Since $\{ k\alpha \} \overrightarrow{OA} + \{ k\beta \} \overrightarrow{OB} = k(\alpha \overrightarrow{OA} + \beta\overrightarrow{OC}) - (\lfloor k\alpha \rfloor \overrightarrow{OA} + \lfloor k\beta \rfloor \overrightarrow{OC})$, the set of points $S = \{ A_k | \overrightarrow{OA_k} = \alpha_k \overrightarrow{OA} + \beta_k \overrightarrow{OC}, 1 \leq k \leq n \}$ is a set of $n$ distinct lattice points inside or on the boundary of $OABC$. If $A_k$ is a point inside or on the boundary of $\triangle OAC$ for a certain $1 \leq k \leq n - 1$, then $\alpha_k + \beta_k \leq 1$ (this is because if $T$ is the intersection point of $OP$ and $AC$ then $\overrightarrow{OT} = \lambda \overrightarrow{OA} + \mu \overrightarrow{OC}$ such that $0 \leq \lambda, \mu \leq 1$ and $\lambda + \mu = 1$). Note that for all real numbers $x \notin \mathbb{Z}, \{ x \} + \{ -x \} = 1$. Hence, $\alpha_k + \alpha_{n - k}= \beta_k + \beta_{n - k} = 1$, that is, $A_{n - k}$ and $A_k$ are symmetric with respect to the midpoint of $AC$, so $A_{n - k}$ is inside or on the boundary of $\triangle ABC$. Among $A_1, A_2, \dots , A_{n - 1}$, the number of points that belong to $\triangle OAC$ and the number of points that belong to $\triangle ABC$ are equal. Together with $A_n = O$, we get $|S \cap \triangle OAC| = |S \cap \triangle ABC| + 1$. Therefore, by pigeonhole, there exists $1 \leq k \leq \Bigl \lfloor \dfrac{n + 1}{2} \Bigr \rfloor,$ such that $A_k$ and $A_{n + 1 - k}$ both belong to $\triangle OAC$. Then $0 \leq \alpha_k + \beta_k \leq 1$ and $0 \leq \alpha_{n + 1 - k} + \beta_{n + 1 - k} \leq 1$. In addition, $P$ belongs to $\triangle ABC$, thus $1 < \alpha + \beta < 2$. Now we have $(\alpha_k + \alpha_{n + 1 - k} - \alpha) + (\beta_k + \beta_{n + 1 - k} - \beta) = (\alpha_k + \beta_k) + (\alpha_{n + 1 - k} + \beta_{n + 1 - k}) - (\alpha + \beta) < 2 - 1 = 1$. Combined with $\alpha_k + \alpha_{n + 1 - k} - \alpha, \beta_k + \beta_{n + 1 - k} - \beta \in \mathbb{Z}$ (since $\{ k\alpha \} + \{ n\alpha - k\alpha + \alpha \} - \alpha = k\alpha + n\alpha - k\alpha + \alpha - \alpha - \lfloor k\alpha \rfloor + \lfloor n\alpha - k\alpha + \alpha \rfloor = n\alpha - \lfloor k\alpha \rfloor + \lfloor n\alpha - k\alpha + \alpha \rfloor \in \mathbb{Z}$), we end up with $(\alpha_k + \alpha_{n + 1 - k} - \alpha) + (\beta_k + \beta_{n + 1 - k} - \beta) \leq 0$. From $\alpha, \alpha_k, \alpha_{n + 1 - k}, \beta, \beta_k, \beta_{n + 1 - k} \in [0, 1)$ and $\alpha_k + \alpha_{n + 1 - k} - \alpha, \beta_k + \beta_{n + 1 - k} - \beta \in \mathbb{Z}$, we have $\alpha_k + \alpha_{n + 1 - k} - \alpha \geq 0$ and $\beta_k + \beta_{n + 1 - k} - \beta \geq 0$. Together with $(\alpha_k + \alpha_{n + 1 - k} - \alpha) + (\beta_k + \beta_{n + 1 - k} - \beta) \leq 0$ we can conclude that $\alpha_k + \alpha_{n + 1 - k} = \alpha$ and $\beta_k + \beta_{n + 1 - k} = \beta$. So take $Q = A_k$ and $R = A_{n + 1 - k}$, which satisfies $\overrightarrow{OP} = \overrightarrow{OQ} + \overrightarrow{OR}$, and we are done. $\blacksquare$