To determine the smallest positive integer value of $m$ for which the equation $mx^{2}-3x+1=0$ has no real solutions, we analyze the discriminant of the quadratic equation. The discriminant ($\Delta$) of a quadratic equation $ax^{2}+bx+c=0$ is given by $\Delta = b^{2}-4ac$. For the equation to have no real solutions, the discriminant must be less than zero ($\Delta < 0$).

Given the equation is $mx^{2}-3x+1=0$, we identify $a = m$, $b = -3$, and $c = 1$. Substituting these values into the formula for the discriminant, we get:
$$\Delta = (-3)^{2}-4(m)(1) = 9-4m$$

For the equation to have no real solutions, we require:
$$9-4m < 0$$

Solving this inequality for $m$ gives:
$$9 < 4m \implies \frac{9}{4} < m$$

Since $m$ must be a positive integer greater than $\frac{9}{4}$, the smallest possible value of $m$ that satisfies this condition is $3$. Therefore, the smallest positive integer value of $m$ for which the equation has no real solutions is $\boxed{3}$.