1. **State the problem:** We are given a natural number $n$ such that $5n - 1 < 3n + 1$. We want to find the least integer value of $m$ such that $3n + 1 < 2n + m$ holds for every such $n$. 2. **Analyze the first inequality:** $$5n - 1 < 3n + 1$$ Simplify by subtracting $3n$ from both sides: $$5n - 3n - 1 < 1$$ $$2n - 1 < 1$$ Add 1 to both sides: $$2n < 2$$ Divide both sides by 2: $$n < 1$$ Since $n$ is a natural number (positive integer), the only natural number less than 1 is none. So no natural number satisfies this inequality. However, if we consider $n$ as a natural number starting from 1, the inequality $5n - 1 < 3n + 1$ is false for all $n \\geq 1$. 3. **Re-examine the problem:** Since no natural number $n$ satisfies $5n - 1 < 3n + 1$, the set of such $n$ is empty. 4. **Interpretation:** If there are no $n$ satisfying the first inequality, then the condition $3n + 1 < 2n + m$ for each such $n$ is vacuously true for any $m$ because there are no $n$ to check. 5. **Conclusion:** The least integer value of $m$ that satisfies the second inequality for all such $n$ is not restricted by any $n$ and can be any integer. The smallest integer is $m = 0$. **Final answer:** $$m = 0$$