4. The highest point of the ellipse obtained by rotating the ellipse $\frac{x^{2}}{2}+y^{2}=1$ counterclockwise by $45^{\circ}$ around the coordinate origin is $\qquad$ .

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4. $\frac{\sqrt{15}}{3}$.

The tangent line to the ellipse at its highest point has a slope of 0. After rotating the ellipse clockwise by $45^{\circ}$ to restore it, the slope of its tangent line is -1.
The point of tangency can be found to be $\left(\frac{2 \sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)$.
Therefore, the required distance is $\frac{\sqrt{15}}{3}$.